GIFT  OF 


ROBBINS'S  NEW 
SOLID  GEOMETRY 


BY 

EDWARD  RUTLEDGE  ROBBINS,  A.B. 

FORMERLY    OF   LAWRENCEVILLE    SCHOOL 


AMERICAN   BOOK   COMPANY 

NEW  YORK  CINCINNATI  CHICAGO 


COPYEIGHT,    1916,   BY 

EDWARD  RUTLEDGE   ROBBINS. 


ROBBINS'8   NEW   SOLID  GEOMETRY. 
W.  P.    I 


FOR    THOSE    WHOSE    PRIVILEGE 
IT    MAY    BE    TO    ACQUIRE    A    KNOWLEDGE    OP 

GEOMETRY 

THIS    VOLUME    HAS    BEEN    WRITTEN 

AND    TO    THE    BOYS    AND    GIRLS    WHO    LEARN    THE    ANCIENT    SCIENCE 

FROM    THESE    PAGES,    AND    WHO    ESTEEM    THE    POWER 

OF    CORRECT    REASONING    THE    MORE 

BECAUSE    OF    THE    LOGIC    OF 

PURE   GEOMETRY 

THIS     VOLUME     IS     DEDICATED 


459960 


PREFACE 

THIS  "  New  Solid  Geometry  "  is  not  only  the  outgrowth  of  the 
author's  long  experience  in  teaching  geometry,  but  has  profited 
further  by  suggestions  from  teachers  who  have  used  Robbins's 
"  Solid  Geometry  "  and  by  the  recommendations  of  the  "  National 
Committee  of  Fifteen."  While  many  new  and  valuable  features 
have  been  added  in  the  reconstruction,  yet  all  the  characteristics 
that  met  with  widespread  favor  in  the  old  book  have  been 
retained. 

Among  the  features  of  the  book  that  make  it  sound  and  teach- 
able may  be  mentioned  the  following : 

1.  The  book  has  been  written  for  the  pupil.     The  objects  sought 
in  the  study  of  Geometry  are  (1)  to  train  the  mind  to  accept 
only  those  statements  as  truth  for  which  convincing  reasons  can 
be  provided,  and  (2)  to  cultivate  a  foresight  that  will  appreci- 
ate both  the  purpose  in  making  a  statement  and  the  process  of 
reasoning  by  which  the  ultimate  truth  is  established.     Thus,  the 
study  of  this  formal  science  should  develop  in  the  pupil  the 
ability  to  pursue  argument  coherently,  and  to  establish  geometric 
truths  in  logical  order.     To  meet  the  requirements  of  the  various 
degrees  of  intellectual  capacity  and  maturity  in  every  class,  the 
reason  for  every  statement  is  riot  printed  in  full  but  is  indicated 
by  a  reference.     The  pupil  who  knows  the  reason  need  not  con- 
sult the  paragraph  cited ;  while  the  pupil  who  does  not  know  it 
may  learn  it  by  the  reference.     It  is  obvious  that  the  greater 
progress  an  individual  makes  in  assimilating  the  subject  and  in 
entering  into  its  spirit,  the  less  need  there  will  be  for  the  printed* 
reference. 

2.  Every  effort  has  been  made  to  stimulate  the  mental  activity 
of  the  pupil.     To  compel  a  young  student,  however,  to  supply  his 
own  demonstrations  frequently  proves  unprofitable  as  well  as 


vi  PREFACE 

arduous,  and  engenders  in  the  learner  a  distaste  for  a  study  in 
which  he  might  otherwise  take  delight.  This  text  does  not  aim 
to  produce  accomplished  geometricians  at  the  completion  of  the 
first  book,  but  to  aid  the  learner  in  his  progress  throughout  the 
volume,  wherever  experience  has  shown  that  he  is  likely  to 
require  assistance.  It  is  designed,  under  good  instruction,  to 
develop  a  clear  conception  of  the  geometric  idea,  and  to  produce 
at  the  end  of  the  course  a  rational  individual  and  a  friend  of  this 
particular  science. 

3.  The  theorems  and  their  demonstrations  —  the  real  subject 
matter  of  Geometry  —  are  introduced  as  early  in  the  study  as 
possible. 

4.  The  simple  fundamental   truths  are  explained  instead  of 
being  formally  demonstrated. 

5.  The  original  exercises  are  distinguished  by  their  abundance, 
their   practical   bearing  upon   the  affairs  of   life,   their  careful 
gradation  and  classification,  and  their  independence.     Every  ex- 
ercise can  be  solved  or  demonstrated  without  the  use  of  any  other 
exercise.     Only  the  truths  in  the  numbered  paragraphs  are  nec- 
essary in  working  originals. 

6.  The  exercises  are  introduced  as  near  as  practicable  to  the 
theorems  to  which  they  apply. 

7.  Emphasis  is  given  to  the  discussion  of  original  constructions. 

8.  The  historical  notes  give  the  pupil  a  knowledge  of  the  devel- 
opment of  the  science  of  geometry  and  add  interest  to  the  study. 

9.  The  attractive  open  page  will  appeal  alike  to  pupils  and 
to  teachers. 

10.  The  Solid  Geometry  formulas  are  grouped  together  at  the 
end  of  the  text,  as  a  ready  means  of  reference. 

The  author  desires  to  extend  his  sincere  thanks  to  those 
friends  and  fellow  teachers  who,  by  suggestion  and  encourage- 
ment, have  inspired  him  in  the  preparation  of  the  text,  as  well 
as  to  Clement  B.  Davis  for  his  original  and  skillful  treatment 
of  the  illustrations. 

EDWARD   R.    ROBBINS. 


CONTENTS 


PAGE 

REFERENCES  TO  "NEW  PLANE   GEOMETRY"    ,       ix 


BOOK  VI.    LINES,  PLANES,  AND  ANGLES  IN  SPACE 

DEFINITIONS 261 

PRELIMINARY  THEOREMS 262 

THEOREMS  AND  DEMONSTRATIONS 264 

ORIGINAL  EXERCISES .        .  287 

DIHEDRAL  ANGLES 290 

ORIGINAL  EXERCISES         . 303 

POLYHEDRAL  ANGLES 305 

ORIGINAL  EXERCISES 312 

BOOK  VII.    POLYHEDRONS 

DEFINITIONS 313 

PRISMS .        .        .  .  314 

PRELIMINARY  THEOREMS .  316 

THEOREMS  AND  DEMONSTRATIONS 317 

ORIGINAL  EXERCISES        .  330 

PYRAMIDS 332 

PRELIMINARY  THEOREMS 333 

THEOREMS  AND  DEMONSTRATIONS 334 

REGULAR  AND  SIMILAR  POLYHEDRONS 353 

ORIGINAL  EXERCISES 360 

BOOK  VIII.     CYLINDERS  AND  CONES 

DEFINITIONS 367 

CYLINDERS 367 

PRELIMINARY  THEOREMS 368 

vii 


viii  CONTENTS 

PAGE 

THEOREMS  AND  DEMONSTRATIONS   . 369 

ORIGINAL  EXERCISES 376 

CONES 377 

PRELIMINARY  THEOREMS 379 

THEOREMS  AND  DEMONSTRATIONS 381 

ORIGINAL  EXERCISES 388 

BOOK  IX.    THE   SPHERE 

DEFINITIONS 893 

PRELIMINARY  THEOREMS 395 

THEOREMS  AND  DEMONSTRATIONS 397 

CONSTRUCTION  PROBLEMS .        .  404 

SPHERICAL  TRIANGLES 408 

PRELIMINARY  THEOREMS 410 

THEOREMS  AND  DEMONSTRATIONS 411 

ORIGINAL  EXERCISES 424 

SPHERICAL  AREAS  AND  VOLUMES 427 

PRELIMINARY  THEOREMS 429 

THEOREMS  AND  DEMONSTRATIONS 431 

ORIGINAL  EXERCISES 443 

SUMMARY  OF  FORMULAS 451 

INDEX  OF  DEFINITIONS 453 


REFERENCES  TO 
ROBBINS'S    "NEW   PLANE   GEOMETRY" 


NOTE.  Many  of  the  theorems  in  the  "  New  Solid  Geometry  "  make  reference 
in  their  proof  to  theorems  and  definitions  in  the  "  New  Plane  Geometry."  These 
references  are  here  collected  for  the  convenience  of  the  pupil. 

16.  One  line  is  perpendicular  to  another  if  they  meet  at  right  angles. 
Either  line  is  perpendicular  to  the  other. 

21.  Parallel  lines  are  straight  lines  that  lie  in  the  same  plane  and  that 
never  meet,  however  far  they  are  extended  in  either  direction. 

23.  A  triangle  is  a  portion  of  a  plane  bounded  by  three  straight 
lines. 

26.  Two  geometric  figures  are  said  to  be  equal  if  they  have  the  same 
size  or  magnitude. 

Two  geometric  figures  are  said  to  be  congruent  if,  when  one  is  super- 
posed upon  the  other,  they  coincide  in  all  respects. 

27.  Homologous  parts  of  congruent  figures  are  equal. 

29.  Symbols.  The  usual  symbols  and  abbreviations  employed  in 
geometry  are  the  following  : 

+  plus. 

—   minus. 

=  equals,  is  equal  to. 

=f=.  does  not  equal. 

^  congruent,  or  is  con- 
gruent to. 

>   is  greater  than. 

<  is  less  than. 

.'.  hence,  therefore,  con- 
sequently. 

_L   perpendicular. 

Ji    perpendiculars. 

O  circle. 


(D  circles. 

Ax. 

axiom. 

/.   angle. 

Hyp. 

hypothesis. 

A  angles. 

comp. 

complementary. 

rt.  /.  right  angle. 

supp. 

supplementary. 

rt.  A  right  angles. 

Const. 

construction. 

A  triangle. 

Cor. 

corollary. 

A  triangles. 

St. 

straight. 

rt.  &  right  triangles. 

rt. 

right. 

||    parallel. 

Def. 

definition. 

Us  parallels. 

alt. 

alternate. 

£7  parallelogram. 

int. 

interior. 

H7  parallelograms. 

ext. 

exterior. 

x      REFERENCES   TO   "NEW   PLANE   GEOMETRY" 

30.  An  axiom  is  a  statement  admitted  without  proof  to  be  true. 

31.  AXIOMS. 

1.  Magnitudes  that  are  equal  to  the  same  thing,  or  to 

equals,  are  equal  to  each  other. 

2.  If  equals  are  added  to,  or  subtracted  from,  equals,  the 

results  are  equal. 

3.  If  equals  are  multiplied  by,  or  divided  by,  equals,  the 

results  are  equal. 
[Doubles  of  equals  are  equal ;  halves  of  equals  are  equal.] 

4.  The  whole  is  equal  to  the  sum  of  all  of  its  parts. 

5.  The  whole  is  greater  than  any  of  its  parts. 

6.  A  magnitude  may  be  replaced  by  its  equal  in  any  process. 

[Briefly  called  "  substitution."] 

7.  If  equals  are  added  to,  or  subtracted  from,  unequals, 

the  results  are  unequal  in  the  same  order. 

8.  If  unequals  are  added  to  unequals  in  the  same  order, 

the  results  are  unequal  in  that  order. 

9.  If  unequals  are  subtracted  from  equals,  the  results  are 

unequal  in  the  opposite  order. 

10.  Doubles  or  halves  of  unequals  are  unequal  in  the  same 

order.     Also,    unequals    multiplied    by    equals    are 
unequal  in  the  same  order. 

11.  If  the  first  of  three  magnitudes  is  greater  than  the 

second,  and  the  second  is  greater  than  the  third,  the 
first  is  greater  than  the  third. 

12.  A  straight  line  is  the  shortest  line  that  can  be  drawn 

between  two  points. 

13.  Only  one  line  can  be  drawn  through  a  point  parallel 

to  a  given  line. 

14.  A  geometrical  figure  may  be  moved  from  one  position 

to  another  without  any  change  in  form  or  magnitude. 


REFERENCES   TO   -NEW  PLAXE   GEOMETRY ~      xi 

39.    Only  one  straight  line  can  be  drawn  between  two  points. 

42.  All  right  angles  are  equal. 

43.  Only  one  perpendicular  to  a  line  can  be  drawn  from  a 
point  in  the  line. 

44.  If  two  adjacent  angles  have  their  exterior  sides  in  a  straight 
line,  they  are  supplementary. 

45.  If  two  adjacent  angles  are  supplementary,  their  exterior 
sides  are  in  the  same  straight  line. 

46.  The  sum  of  all  the  angles  on  one  side  of  a  straight  line  at  a 
point  equals  two  right  angles. 

47.  The  sum  of  all  the  angles  about  a  point  in  a  plane  is  equal 
to  four  right  angles. 

49.  Angles  that  have  the  same  supplement  are  equal.  Or, 
supplements  of  the  same  angle,  or  of  equal  angles,  are  equal. 

51.  If  two  straight  lines  intersect,  the  vertical  angles  are  equal. 

52.  Two  triangles  are  congruent  if  two  sides  and  the  included 
angle  of  one  are  equal  respectively  to  two  sides  and  the  included 
angle  of  the  other. 

53.  Two  right  triangles  are  congruent  if  two  legs  of  one  are 
equal  respectively  -to  two  legs  of  the  other. 

54.  Only  one  perpendicular  can  be  drawn 'to  a  line  from  an 
external  point. 

55.  The  angles  opposite  the  equal  sides  of  an  isosceles  triangle 
are  equal. 

62.  Two  lines  in  the  same  plane  and  perpendicular  to  the  same 
line  are  parallel. 

64.  If  a  line  is  perpendicular  to  one  of  two  parallels,  it  is  per- 
pendicular to  the  other  also. 

66.  If   a   transversal   intersects   two   parallels,    the   alternate 
interior  angles  are  equal. 

67.  If  a  transversal  intersects  two  parallels,  the  corresponding 
angles  are  equal. 


xii    REFERENCES   TO   "NEW   PLANE   GEOMETRY" 

76.  Two  triangles  are  congruent  if  a  side  and  the  two  angles 
adjoining  it  in  the  one  are  equal  respectively  to  a  side  and  the 
two  angles  adjoining  it  in  the  other. 

77.  Two  right  triangles  are  congruent  if  a  leg  and  the  adjoin- 
ing acute  angle  of  one  are  equal  respectively  to  a  leg  and  the  ad- 
joining acute  angle  of  the  other. 

78.  Two  triangles  are  congruent,  if  the  three  sides  of  one  are 
equal  respectively  to  the  three  sides  of  the  other. 

80.  Any  point  in  the  perpendicular  bisector  of  a  line  is  equally 
distant  from  the  extremities  of  the  line. 

81.  Any  point  not  in  the  perpendicular  bisector  of  a  line  is  not 
equally  distant  from  the  extremities  of  the  line. 

82.  If  a  point  is  equally  distant  from  the  extremities  of  a  line, 
it  is  in  the  perpendicular  bisector  of  the  line. 

83.  Two  points  each  equally  distant  from  the  extremities  of  a 
line  determine  the  perpendicular  bisector  of  the  line. 

84.  Two  right  triangles  are  congruent  if  the  hypotenuse  and  a 
leg  of  one  are  equal  respectively  to  the  hypotenuse  and  a  leg  of 
the  other. 

87.  The  perpendicular  is  the  shortest  line  that  can  be  drawn 
from  a  point  to  a  straight  line. 

88.  If  from  any  point  in  a  perpendicular  to  a  line  two  oblique 
lines  are  drawn, 

I.  Oblique  lines  cutting  off  equal  distances  from  the  foot  of 
the  perpendicular  are  equal. 

II.  Equal  oblique  lines  cut  off  equal  distances. 

III.  Oblique  lines  cutting  off  unequal  distances  are  unequal, 
and  that  one  which  cuts  off  the  greater  distance  is  the  greater. 

90.  The  method  of  exclusion  consists  in  making  all  possible  supposi- 
tions, leaving  the  probable  one  last,  and  then  proving  all  these  supposi- 
tions impossible,  except  the  last,  which  must  necessarily  be  true. 


REFERENCES   TO   "NEW   PLANE   GEOMETRY"     xiii 

The  method  of  proving  the  individual  steps  is  called  reductio  ad 
absurdum  (reduction  to  an  absurd  or  impossible  conclusion).  This 
method  consists  in  assuming  as  false  the  truth  to  be  proved  and  then 
showing  that  this  assumption  leads  to  a  conclusion  altogether  contrary 
to  known  truth  or  the  given  hypothesis. 

92.  If  two  triangles  have  two  sides  of  one  equal  to  two  sides 
of  the  other,  but  the  third  side  of  the  first  greater  than  the  third 
side  of  the  second,  the  included  angle  of  the  first  is  greater  than 
the  included  angle  of  the  second. 

94.  Every  point  in  the  bisector  of  an  angle  is  equally  distant 
from  the  sides  of  the  angle. 

104.  The  sum  of  the  angles  of  any  triangle  is  two  right  angles ; 
that  is,  180°. 

109.   Each  angle  of  an  equiangular  triangle  is  60°. 

114.  •  If  two  angles  of  a  triangle  are  equal,  the  triangle  is  isosceles. 

120.  A  parallelogram  is  a  quadrilateral  having  its  opposite  sides 
parallel. 

124.   The  opposite  sides  of  a  parallelogram  are  equal. 
126.   The  diagonal  of  a  parallelogram  divides  it  into  two  con- 
gruent triangles. 

128.  If  the  opposite  sides  of  a  quadrilateral  are  equal,  the  figure 
is  a  parallelogram. 

129.  If  two  sides  of  a  quadrilateral  are  equal  and  parallel,  the 
figure  is  a  parallelogram. 

133.  Two  parallelograms  are  congruent  if  two  sides  and  the 
included  angle  of  one  are  equal  respectively  to  two  sides  and  the 
included  angle  of  the  other. 

134.  Two  rectangles  are  congruent  if  the  base  and  altitude 
of  one  are  equal  respectively  to  the  base  and  altitude  of  the  other. 

136.  The  line  joining  the  midpoints  of  two  sides  of  a  triangle  is 
parallel  to  the  third  side  and  equal  to  half  of  it. 

138.  The  line  bisecting  one  leg  of  a  trapezoid  and  parallel  to  the 
base  bisects  the  other  leg,  is  the  median,  and  is  equal  to  half  the 
sum  of  the  bases. 


xiv     REFERENCES   TO   "NEW   PLANE   GEOMETRY" 

139.  The  median  of  a  trapezoid  is  parallel  to  the  bases  and  equal 
to  half  their  sum. 

143.  The  three  medians  of  a  triangle  meet  in  a  point  which  is 
two  thirds  the  distance  from  any  vertex  to  the  midpoint  of  the 
opposite  side. 

145.  The  number  of  sides  of  a  polygon  is  the  same  as  the  number  of 
its  vertices  or  the  number  of  its  angles. 

160.  Two  polygons  are  congruent  if  they  are  mutually  equiangular 
and  their  homologous  sides  are  equal. 

153.  The  sum  of  the  interior  angles  of  an  n-gon  is  equal  to 
(n-2)  times  180°. 

155.   Each  angle  of  an  equiangular  n-gon  =  (-n~*>  18°  . 

n 

167.  If  three  angles  of  a  quadrilateral  are  right  angles,  the  figure 
is  a  rectangle. 

168.  If  the  sides  of  a  polygon  are  produced,  in  order,  one  at 
each  vertex,  the  sum  of  the  exterior  angles  of  the  polygon  equals 
four  right  angles,  that  is,  360°. 

179.  A  circle  is  a  plane  curve  all  points  of  which  are  equally  distant 
from  a  point  in  the  plane,  called  the  center. 

180.  The  length  of  the  circle  is  called  the  circumference. 
183.   Equal  circles  are  circles  having  equal  radii. 

187.  All  radii  of  the  same  circle  are  equal. 

188.  All  radii  of  equal  circles  are  equal. 

190.  All  diameters  of  the  same  or  of  equal  circles  are  equal. 

191.  The  diameter  of  a  circle  bisects  the  circle. 

193.  In  the  same  circle  (or  in  equal  circles)  equal  central  angles 
intercept  equal  arcs. 

196.  In  the  same  circle  (or  in  equal  circles)  equal  chords  sub- 
tend equal  arcs. 

197.  In  the  same  circle  (or  in  equal  circles)  equal  arcs  are  sub- 
tended by  equal  chords. 


REFERENCES   TO   "NEW  PLANE   GEOMETRY"     xv 

202.  The  line  perpendicular  to  a  radius  at  its  extremity  is  tangent 
to  the  circle. 

203.  If  a  line  is  tangent  to  a  circle,  the  radius  drawn  to  the  point 
of  contact  is  perpendicular  to  the  tangent. 

208.  In  the  same  circle  (or  in  equal  circles)  equal  chords  are 
equally  distant  from  the  center. 

209.  In  the  same  circle  (or  in  equal  circles)  chords  which  are 
equally  distant  from  the  center  are  equal. 

210.  In  the  same  circle  (or  in  equal  circles)  if  two  chords  are  un- 
equal, the  greater  chord  is  at  the  less  distance  from  the  center. 

211.  In  the  same  circle  (or  in  equal  circles)  if  two  chords  are  un- 
equally distant  from  the  center,  the  chord  at  the  less  distance  is  the 
greater. 

214.  One  circle  and  only  one  can  be  drawn  through  the  vertices 
of  a  triangle. 

219.  If  two  circles  intersect,  the  line  joining  their  centers  is  the 
perpendicular  bisector  of  their  common  chord. 

224.  To  measure  a  quantity  is  to  find  the  number  of  times  it  contains 
another  quantity  of  the  same  kind,  called  the  unit.     This  number  is  the 
ratio  of  the  quantity  to  the  unit. 

225.  Two  quantities  are  called  commensurable  if  there  exists  a  com- 
mon  unit  of   measure  which  is  contained  in  each   a  whole  (integral) 
number  of  times. 

Two  quantities  are  called  incommensurable  if  there  does  not  exist 
a  common  unit  of  measure  which  is  contained  in  each  a  whole  number 
of  times. 

227.  The  limit  of  a  variable  is  a  constant,  to  which  the  variable  can- 
not be  equal,  but  from  which  the  variable  can  be  made  to  differ  by  less 
than  any  mentionable  quantity. 

229.  If  two  variables  are  always  equal  and  each  approaches  a 
limit,  their  limits  are  equal. 

232.  A  central  angle  is  measured  by  its  intercepted  arc. 

233.  A  central  right  angle  intercepts  a  quadrant  of  arc. 

234.  A  right  angle  is  measured  by  half  a  semicircle,  that  is,  by  a 
quadrant. 


xvi     REFERENCES   TO   "NEW   PLANE   GEOMETRY" 

240.   All  angles  inscribed  in  a  semicircle  are  right  angles. 

246.  The  locus  of  a  point  is  the  series  of  positions  the  point,  must 
occupy  in  order  that  it  may  satisfy  a  given  condition.  It  is  the  path  of 
a  point  whose  positions  are  limited  or  defined  by  a  given  condition,  or 
given  conditions. 

280.  In  a  proportion  the  product  of  the  extremes  is  equal  to  the 
product  of  the  means. 

281.  If  the  product  of  two  quantities  is  equal  to  the  product  of 
two  others,  one  pair  may  be  made  the  extremes  of  a  proportion  and 
the  other  pair  the  means. 

282.  In  any  proportion  the  terms  are  also  in  proportion  by  alter- 
nation (that  is,  the  first  term  is  to  the  third  as  the  second  is  to  the  fourth). 

284.  In  any  proportion  the  terms  are  also  in  proportion  by  compo- 
sition (that  is,  the  sum  of  the  first  two  terms  is  to  the  first,  or  the  second, 
as  the  sum  of  the  last  two  terms  is  to  the  third,  or  the  fourth). 

286.  In  any  proportion  the  terms  are  also  in  proportion  by  division 

(that  is,  the  difference  between  the  first  two  terms  is  to  the  first,  or  the 
second,  as  the  difference  between  the  last  two  terms  is  to  the  third,  or 
the  fourth). 

287.  In  any  proportion,  like  powers  of  the  terms  are  in  propor- 
tion, and  like  roots  of  the  terms  are  in  proportion. 

288.  In  two  or  more  proportions  the  products  of  the  correspond- 
ing terms  are  in  proportion. 

289.  A  mean  proportional  is  equal  to  the  square  root  of  the 
product  of  the  extremes. 

291.  In  a  series  of  equal  ratios,  the  sum  of  all  the  antecedents  is 
to  the  sum  of  all  the  consequents  as  any  antecedent  is  to  its  conse- 
quent. 

293.  A  line  parallel  to  one  side  of  a  triangle  divides  the  other 
sides  into  proportional  segments. 

294.  If  a  line  parallel  to  one  side  of  a  triangle  intersects  the 
other  sides,  it  divides  these  sides  proportionally. 

301.  Similar  polygons  are  polygons  that  are  mutually  equiangular 
and  whose  homologous  sides  are  proportional. 


REFERENCES   TO   "NEW  PLANE   GEOMETRY"     xvii 

304.  Two  right  triangles  are  similar  if  an  a'cute  angle  of  one  is 
equal  to  an  acute  angle  of  the  other. 

305.  If  a  line  parallel  to  one  side  of  a  triangle  intersects  the 
other  sides,  the  triangle  formed  is  similar  to  the  original  triangle. 

308.  If  two  triangles  have  their  homologous  sides  proportional, 
they  are  similar. 

310.  If  two  triangles  have  their  homologous  sides  perpendicular, 
they  are  similar. 

313.   In  similar  figures  homologous  sides  are  proportional. 

318.  If  two  polygons  are  similar,  they  may  be  decomposed  into 
the  same  number  of  triangles  similar  each  to  each  and  similarly 
placed. 

331.  If  in  a  right  triangle  a  perpendicular  is  drawn  from  the 
vertex  of  the  right  angle  upon  the  hypotenuse, 

I.  The  triangles  formed  are  similar  to  the  given  triangle  and 
similar  to  each  other. 

II.  The  perpendicular  is  a  mean  proportional  between  the  seg- 
ments of  the  hypotenuse. 

333.  The  square  of  a  leg  of  a  right  triangle  is  equal  to  the 
product  of  the  hypotenuse  and  the  projection  of  this  leg  upon  the 
hypotenuse. 

334.  The  sum  of  the  squares  of  the  legs  of  a  right  triangle  is 
equal  to  the  square  of  the  hypotenuse. 

357.  The  area  of  a  rectangle  is  equal  to  the  product  of  its  base 
by  its  altitude. 

359.  The  area  of  a  parallelogram  is  equal  to  the  product  of  its 
base  by  its  altitude. 

360.  All  parallelograms  having  equal  bases  and  equal  altitudes 
are  equal  in  area. 

364.  The  area  of  a  triangle  is  equal  to  half  the  product  of  its 
base  by  its  altitude. 

366.  All  triangles  having  equal  bases  and  equal  altitudes  are 
equal  in  area. 

ROBBINS'8   NEW   SOLID    GEOM. —  2 


xviii     REFERENCES  TO   "NEW   PLANE   GEOMETRY" 

368.  Two  triangles  having  equal  altitudes  are  to  each  other  as 
their  bases. 

372.  The  area  of  a  trapezoid  is  equal  to  half  the  product  of  the 
altitude  by  the  sum  of  the  bases. 

374.  If  two  triangles  have  an  angle  of  one  equal  to  an  angle  of 
the  other,  they  are  to  each  other  as  the  products  of  the  sides  includ- 
ing the  equal  angles. 

376.  Two  similar  polygons  are  to  each  other  as  the  squares  of 
any  two  homologous  sides. 

395.   To  construct  a  triangle  equal  to  a  given  polygon. 

422.  If  the  number  of  sides  of  an  inscribed  regular  polygon  is 
indefinitely  increased,  the  apothem  approaches  the  radius  as  a 
limit. 

NOTE,  page  232.  It  is  evident  that  if  the  difference  between  two  variables 
approaches  zero,  either  ' 

(1)  one  is  approaching  the  other  as  a  limit,  or 

(2)  both  are  approaching  some  third  quantity  as  their  limit. 

424.  If  the  number  of  sides  of  an  inscribed  regular  polygon  and 
of  a  circumscribed  regular  polygon  is  indefinitely  increased, 

I.   The  perimeter  of    each   polygon  approaches    the  circum- 
ference of  the  circle  as  a  limit. 

II.  The  area  of  each  polygon  approaches  the  area  of  the  circle 
as  a  limit. 

428.   Let  C  =  circumference  and  R  =  radius.     Then,  C  =  2-rrR. 

430.  Let  S  =  area  of  O,  C  =  its  circumference,  and  R  =  its  radius. 
Then,  S  = 


SOLID   GEOMETRY 

BOOK  VI 

LINES,  PLANES,   AND   ANGLES  IN  SPACE 

465.  A  solid  is  any  limited  portion  of  space.     The  bound- 
aries of  a  solid  are  surfaces. 

A  plane  is  a  surface  in  which,  if  any  two  points  are  taken, 
the  straight  line  connecting  them 
lies  wholly  in  that  surface. 

Solid  geometry  is  a  science 
that  treats  of  magnitudes,  the 
parts  of  which  are  not  all  in  the 
same  plane. 

466.  The  intersection  of   two  surfaces  is  the  line,  or  the 
lines,  all  of  whose  points  lie  in  both  surfaces.     The  inter- 
section of  a  line  and  a  surface  is  the  point,  or  the  points, 
common  to  both  the  line  and  the  surface.     The  foot  of  a 
line  intersecting  a  plane  is  their  point  of  intersection. 

467.  A  straight  line  is  perpendicular  to  a  plane  if  the  line 
is  perpendicular  to  every  straight  line  in  the  plane  drawn 
through  its  foot. 

A  normal  is  a  straight  line  perpendicular  to  a  plane. 

468.  A  straight  line  is  parallel  to  a  plane  if  the  line  and  the 
plane  never  meet,  when  indefinitely  extended.     A  straight 
line  is  oblique  to  a  plane  if  it  is  neither  perpendicular  nor 
parallel  to  the  plane.     Two  planes  are  parallel  if  they  never 
meet  when  indefinitely  extended. 

261 


262 


VI.     SOLID   GEOMETRY 


469.  The  projection  of  a  point  upon  a  plane  is  the  foot  of 
the  perpendicular  from  the  point  to  the  plane. 

The  projection  of  a  line  upon  a  plane  is  the  line  formed  by 
the  projections  of  all  the  points  of  the  given  line. 

470.  A  plane  is  determined  if  its  position  is  fixed  and  if 
that  position  can  be  occupied  by  only  one  plane. 

PRELIMINARY   THEOREMS 

471.  If  two  points  of  a  straight  line  are  in  a  plane,  the  whole 
line  is  in  the  plane.  (Def.  465.) 

472.  A  straight  line  can  intersect  a  plane  in  only  one  point. 

(471.) 

473.  If  a  line  is  perpendicular  to  a  plane,  it  is  perpendicular 
to  every  line  in  the  plane  drawn  through  its  foot.  (467.) 

474.  Through  one  straight  line  any  number  of  planes  may 
be  passed. 


Because,  if  we  consider  a  plane  containing  a  line  AB  to 
revolve  about  AB,  it  may  occupy  an  indefinitely  great  number 
of  positions.  Each  of  these  will  be  a  different  plane  con- 
taining AB. 

475.  Through  a  fixed  straight  line  and  an  external  point  a 
plane  can  be  passed. 

Because,  if  we  pass  a  plane  containing  this  line  AB,  it  may 
be  revolved  about  AB  until  it  contains  the  given  point. 

476.  A  straight  line  and  an  external  point  determine  a  plane. 

(See  475,  470.) 


LINES    AND   PLANES  263 

477.  Three  points  not  in  a  straight  line  determine  a  plane. 

Because  two  of  the  points  may  be  joined  by  a  line ;  then 
this  line  and  the  third  point  determine  a  plane.  (See  476.) 

478.  Two  parallel  lines  determine  a  plane.  (21,  476.) 

479.  Two  intersecting  straight  lines  determine  a  plane. 
Because  one  of  these  lines  and  a  point  in  the  second  line 

determine  a  plane  ;  and  this  plane  contains  the  second  line. 

(476,  471.) 

480.  If  two  planes  are  parallel,  no  line  in  the  one  can  meet 
any  line  in  the  other.  (Def.  468.) 

NOTE.  A  plane  is  represented  to  the  eye  by  a  quadrilateral.  In  some 
positions  it  appears  to  be  a  parallelogram,  and  in  others,  a  trapezoid. 
The  eye,  however,  must  be  aided  by  the  imagination  in  really  under- 
standing the  diagrams  of  solid  geometry.  Thus,  in  the  adjoining  figure, 
the  line  CN  is  perpendicular  to  the  plane  FR, 
and  to  every  line  in  FR  drawn  through  N. 
Consider  several  lines  drawn  through  a  point 
on  the  floor,  and  a  cane,  CN,  occupying  a  verti- 
cal position,  so  that  it  is  perpendicular  to  all 
these  lines.  Then  every  angle  CNX  is  a  right 
angle,  though  to  the  unskilled  eye  they  do  not 
all  appear  to  be  right  angles  in  the  diagram. 
The  object  of  all  geometrical  diagrams  is  that 
the  eye  may  assist  the  rnind  in  grasping  truths 
or  in  developing  logical  demonstrations,  and 

the  student  should  thoroughly  examine  every  figure  until  he  completely 
understands  the  relative  positions  of  its  parts.  A  photograph,  like  a 
geometrical  diagram,  represents  three  dimensions  in  a  plane,  and  we 
should  be  as  familiar  with  the  significance  of  one  as  with  the  other. 

When,  during  the  process  of  a  demonstration  or  elsewhere,  it  becomes 
necessary  to  employ  a  plane  not  already  indicated,  it  is  customary  to  pass 
such  a  plane,  or  to  conceive  it  constructed. 


Ex.  1.  How  many  planes  can  be  passed  through  two  points?  through 
three  points  in  the  same  straight  line  ? 

Ex.  2.  Hold  two  pencils  (representing  lines)  so  that  a  plane  can  be 
passed  containing  both;  so  that  no  plane  can  be  passed  containing  both. 


264  '  BOOK   VI.     SOLID   GEOMETRY 

THEOREMS  AND  DEMONSTRATIONS 

POINTS,  LINES,  AND  PLANES 
PROPOSITION  I.     THEOREM 

481.   If  two  planes  intersect,  their  intersection  is  a  straight 
line. 


Given :  Intersecting  planes  MN  and  RS. 

To  Prove :  Their  intersection  is  a  straight  line. 

Proof :  Suppose  A  and  B  are  two  points  common  to  both 
planes. 

Draw  straight  line  AB. 

Now  AB  is  in  plane  RS  (471). 

And  AB  is  in  plane  MN  (?). 

That  is,  AB  is  common  to  both  planes. 

Again,  if  there  were  a  point  outside  of  AB  in  both  planes, 
these  planes  would  coincide  (476). 

Hence  AB  contains  all  points  common  to  the  two  given 

Planes*  .-.  AB  is  the  intersection  (466). 

That  is,  the  intersection  of  the  two  planes  is  a  straight 
line.  Q.E.D. 

Ex.  1.     Do  any  two  planes  intersect?     Explain. 

Ex.  2.     What  is  meant  by  the  statement  "  Two  planes  determine  a 
line  "?    Is  this  universally  true? 

Ex.  3.     What  kind  of  lines  are  the  folds  in  your  letter  paper?  in  a 
pamphlet?  in  the  edges  of  a  box  or  a  brick?    Explain. 


LINES   AND   PLANES 


265 


PROPOSITION  II.     THEOREM 

482.   If  two  straight  lines  are  parallel,  a  plane  containing 
one,  and  only  one,  is  parallel  to  the  other  line. 

A 


Given :  II  lines  AB  and  CD ;  plane  MN  containing  CD. 

To  Prove :  plane  MN  II  to  line  AB. 

Proof:      AB  and  CD  are  in  the  same  plane  AD 

Plane  AD  intersects  plane  MN  in  CD  (481). 

Now  as  AB  cannot  meet  MN  in  CD(  hyp.)  it  can  never  meet  MN. 

.-.  AB  is  II  to  MN  (468).    Q.E.D. 

PROPOSITION  III.     THEOREM 

483.  If  a  straight  line  is  parallel  to  a  plane,  and  another 
plane  containing  this  line  intersects  the  given  plane,  the  inter- 
section is  parallel  to  the  given  line. 

C 


M 


B 


Given :  AB  II  to  MN ;  plane  AD  containing  AB  and  inter- 
secting plane  MN  in  CD. 
To  Prove :  AB  II  to  CD. 

Proof:  AB  and  CD  are  in  the  same  plane  AD  (Hyp.). 

Now  a,sAB  cannot  meet  Jfjr(hyp.)it  can  never  meet  CD  in  MN. 

.'.  AB  is  II  to  CD  (21).      Q.E.D. 


266 


BOOK   VI.     SOLID   GEOMETRY 


PROPOSITION  IV.     THEOREM 

484.  The  intersections  of  two  parallel  planes  by  a  third 
plane  are  parallel  lines. 


Given :  if  planes  AB  and  CD  cut  by  plane  ES  in  lines  LM 
and  PQ. 


To  Prove :  LM  II  to  PQ. 

Proof :  LM  and  PQ  are  in  the  same  plane  BS 
Also  LM  and  PQ  can  never  meet 

.-.  LM  is  II  to  PQ 


(Hyp.)- 

(480). 

(21), 

Q.E.D. 


Ex.  1.  Hold  a  pencil  parallel  to  the  blackboard,  so  that  its  shadow 
falls  on  the  blackboard.  Is  this  shadow  parallel  to  the  pencil  ?  Why  ? 

Ex.  2.  Can  a  plane  intersect  two  planes  that  are  not  parallel  so  that 
the  intersections  are  parallel?  Illustrate  your  answer  by  passing  a 
plane  across  the  room  so  that  it  cuts  the  end  and  a  side  of  the  room. 
How  must  this  plane  be  passed  so  that  the  intersections  are  parallel 
lines? 

Ex.  3.  Draw  a  line  on  the  blackboard  that  will  never  meet  the  plane 
of  the  ceiling.  How  was  it  drawn  ? 

Ex.  4.  Draw  a  line  on  the  blackboard  and  one  on  the  floor  that  will 
meet  if  extended.  How  must  these  lines  be  drawn  ? 

Ex.  6.  When  will  a  line  on  the  ceiling  be  parallel  to  a  line  on  the 
floor? 


LINES   AND   PLANES 
PROPOSITION  V.     THEOREM 


267 


485.   A  straight  line  perpendicular  to  each  of  two  straight 
lines  at  their  intersection  is  perpendicular  to  the  plane  of  the 


lines. 


M 


Given:  -4.FJ.to  BF and  CF  at  F-,  plane  MN  containing  BF 
and  CF. 

To  Prove :  AF  JL  to  plane  MN. 

Proof:  In   plane  MN  draw  BC\  draw  also   DF  from  F  to 
any  point,  D,  in  BC. 

Prolong  AF  to  X,  making  FX  =  to  AF,  and  draw  AB,  AD, 
AC,  BX,  DX,  CX. 

BF  and  CF  are  _L  bisectors  of.  AX   (Hyp.  and  Const.). 


In        A  ABC  and  BXC,  AB  =  BX  and  AC  =  CX 

BC=BC 

.-.  A  ABC  ^  A  BXC 
Also  in  A  ^UD  and  £JTD,  £  ABC  =  /.  CBX 

BD  =  BD 

And  AB  —  BX 

.'.  AABD  ^ 

.'.  AD  =  DX 
DF  is  -L  to  AX 


Hence 

That  is, 


(80). 

(?)• 
(78). 
(27). 

(?)• 
CO- 
(52). 

(?)• 

(83). 

AF  is  J_  to  all  lines  in  MN  through  F. 
AF  is  ±  to  plane  Jf^r  (467). 

Q.E.D. 


268 


BOOK   VI.     SOLID   GEOMETRY 


PROPOSITION  VI.     THEOREM 

486.  All  straight  lines  perpendicular  to  a  line  at  one  point 
are  in  one  plane,  which  is  perpendicular  to  this  line  at  this 
point. 

A 


N 


Given :  AB  _L  to  BC,  .BD,  BE,  etc. ;  plane  MN  containing  BC 
and  BD. 

To  Prove :   BE  is  in  the  plane  MN  and  MN  is  -L  to  AB  at  B. 

Proof:   Pass  plane  AE  containing  AB  and  BE,  and  inter- 
secting plane  MN  in  line  BX. 

Now  AB  is  -L  to  plane  MN  (485). 

That  is,  plane  MN  is  J_  to  AB 

.-.  AB  is  -L  to  BX  (473). 

But  AB  is  -L  to  BE  (Hyp.)- 

That  is,  BX  and  BE  are  both  in  plane  AX  and  J-  to  AB  at  B. 

.  •.  BX  and  BE  coincide  (43). 

That  is,  BE  is  in  plane  MN.  Q.E.D. 


Ex.  1.  Can  a  line  be  perpendicular  to  two  other  lines  if  these  two  do 
not  intersect  ?  Illustrate. 

Ex.  2.  If  two  lines  are  perpendicular  to  a  third  line  are  they  neces- 
sarily in  the  same  plane?  Are  the  two  lines  necessarily  parallel?  Illus- 
trate. 

Ex.  3-  Give  a  reason  why  the  "  corners  "  of  a  building  are  perpen- 
dicular to  the  horizontal  plane  of  a  level  street. 


LINES   AND  PLANES  269 

487.  COROLLARY.     Through  a  point  in  a   straight  line  one 
plane  can  be  passed  perpendicular  to  the  line,  and  only  one. 

(486.) 

PROPOSITION  VII.     THEOREM 

488.  Through  an  external  point  one  plane  can  be  passed 
perpendicular  to  a  given  straight  line,  and  only  one. 


IB 

Given :  The  line  AB  and  point  P  outside  of  AB. 

To  Prove :  Through  P,  one  plane  can  be  passed  J_  to  AB, 
and  only  one. 

Proof:  I.     Draw  from  P,  PC  _L  to  AB,  and  at  C  draw  CX, 
another  line  J_  to  AB. 

PC  and  CX  determine  a  plane  MN  (479). 

Plane  MN  contains  P  and  is  -L  to  AB  (485). 

II.     Only  one  line  J_  to  AB  can  be  drawn  from  P       (54). 

And  only  one  plane  _L  to  AB  can  be  passed  at  C         (487). 

That  is,  MN  is  the  only  plane  _L  to  AB  that  can  be  passed 

through  P.  Q.E.D. 

Ex.  1.     Considered  as  lines,  why  are  the  spokes  of  a  cart  wheel  per- 
pendicular to  the  axle? 

Ex.  2.     As  the  hand  of  a  clock  revolves,  what  may  it  be  said  to  describe, 
if  it  is  considered  of  indefinite  length?    Why? 

Ex.  3.     Illustrate  Proposition  VI  by  revolving  a  carpenter's  square, 
holding  one  straight  edge  against  the  wall  or  the  floor. 


2TO 


BOOK  VI.     SOLID   GEOMETRY 


PROPOSITION  VIII.     THEOREM 

489.   Two  planes  perpendicular  to  the  same  straight  line  are 
parallel. 


B 


Given :   Planes  MN  and  OP  J_  to  AB. 
To  Prove :    Plane  MN  \\  to  plane  OP. 

Proof:  If  the  planes  MN  and  OP  are  not  ||,  they  will  meet 
when  sufficiently  extended  (Def.  468). 

Then  there  would  be  two  planes  from  the  same  point  _L  to 
AB  (-L  by  hyp.). 

But  this  is  impossible  (488). 

.  *.  the  planes  never  meet  and  are  parallel  (468).  Q.E.D. 

PROPOSITION  IX.     THEOREM 

490.  At  a  given  point  in  a  plane  one  line  can  be  drawn 
perpendicular  to  the  given  plane,  and  only  one. 


N 


M 


Given :  Plane  MN  and  point  P  within  it. 


LINES   AND  PLANES  271 

To  Prove :  One  line  can  be  drawn  _L  to  plane  'MN  at  P,  and 
only  one. 

Proof:    I.    In  plane  MN  draw  any  line  AB,  through  P. 
Suppose  plane  CD   is  passed  _L  to  AB  at  P,  meeting   the 
plane  MN  in  CE. 

In  plane  CD  draw  PE  J_  to  CE,  from  P. 

Now  AB  is  J-  to  plane  CD  (Const.). 

.-.  AB  is  _L  to  PR  (473). 

PE  is  _L  to  CE  (Const.). 

.-.  PE  is  _L  to  plane  MN         (485).     Q.E.D. 

II.     Suppose  another  line  PX  to  be  _L  to  plane  MN  at  P. 

Then            PX  and  PE  determine  a  plane  CD  (479). 

And  plane  CD  intersects  plane  MN  in  line  CE  (481). 

Then  PX  and  PE  would  both  be  -L  to  CE  at  P  (473). 

But                            this  is  impossible  (43). 
That   is,  PX  and  PE  coincide  and  PE   is  the  only  J_  to 

plane  MN  at  P.  Q.E.D. 

Ex.  1.  How  many  positions  can  a  flagpole  occupy  without  being  per- 
fectly erect?  How  many  positions  may  it  assume  and  be  perfectly 
erect? 

Ex.  2.  Name  all  the  right  angles  at  P,  in  figure  of  490,  and  tell 
why  each  is  a  right  angle. 

Ex.  3.  What  information  can  the  mason  or  the  surveyor  obtain  from 
a  plumb  bob?  Does  he  obtain  this  information  when  the  bob  is  swing- 
ing or  when  it  is  stationary  ? 

Ex.  4.  In  transplanting  a  tree  to  a  horizontal  lawn  a  gardener  may 
use  a  carpenter's  square  to  make  certain  that  the  tree  is  perpendicular  to 
the  lawn.  In  how  many  different  positions  must  he  place  the  square 
against  the  tree  to  ascertain  its  erectness?  Why? 

Ex.  5.  In  the  diagram  of  490,  if  PX  is  in  plane  CD,  is  it  perpen- 
dicular to  CE  ?  '  Why  ? 

Ex.  6.  In  the  same  diagram,  if  PX  is  perpendicular  to  CE,  is  it  in 
plane  GDI  Why? 

Ex.  7.  How  many  planes  are  determined  by  four  random  fixed 
points  (that  is,  not  all  in  one  plane)  ? 


272 


BOOK  VI.     SOLID   GEOMETRY 


PROPOSITION  X.     THEOREM 

491.   Through  a  given  external  point  one  line  can  be  drawn 
perpendicular  to  a  given  plane,  and  only  one. 

G 


Given  :  Plane  MN  and  point  P  outside  of  it. 

To  Prove  :  One  line  can  be  drawn  through  P  _L  to  plane  MN, 
and  only  one. 

Proof  :  I.     In  plane  MN  draw  any  line  AB. 
Suppose  a  plane  GH  is  passed  through  P  _L  to  AB,  meeting 
plane  MN  in  KC,  and  AB  at  C. 

In  plane  GH  draw  PR  J_  to  KG  and  prolong  PR  to  X,  mak- 
ing RX  =  PR. 

Draw  RD  to  any  point  in  AB,  except  C. 

Draw  PC,  PD,  CX,  DX. 

Now  RC  is  .1  to  PX  at  its  midpoint  (Const.). 

Also  AB  is  J_  to  plane  GH  (Const.). 

Hence  A  DCP  and  DCX  are  rt.  A  (473). 

In  rt.  A  DCP  and  DCX,  DC  =  DC  (?). 

PC=CX  (80). 

.*.  A  DCP^A  DCX  (53). 


.-.  JH>  is  JL  to  PX  (83). 

That  is,      PR  is  -L  to  RC  and  RZ>,  in  plane  MN. 

.-.  PR  is  J_  to  plane  JfJV  from  P       (485).     Q.E.D. 

II.    Suppose  there  is  another  line  PL,  _L.to  plane  MN  from  P. 

Then  PR  and  PL  determine  a  plane  GH         (479). 


LINES  AND  PLANES  273 

This  plane  intersects  plane  MN  in  KC          (481). 

PR  and  PL  would  then  both  be  J_  to  KG        (473). 

But  this  is  impossible  (54). 

That  is,  PR  and  PL  coincide,  and  therefore  PR  is  the  only 

line  J-  to  plane  MN  from  P.  Q.E.D. 


Ex.     Name  all  the  right  angles  at  C  in  the  figure  of  491. 

492.  COROLLARY.     If  a  plane  is  perpendicular  to  a  line  in 
another  plane,  any  line  in  the  first  plane  perpendicular  to  the 
intersection  of  the  planes  is  perpendicular  to  the  second  plane. 

Proof :  Identical  with  the  proof  of  491,  I. 

PROPOSITION  XI.     THEOREM 

493.  If  a  plane  is  perpendicular  to  one  of  two  parallel  lines, 
it  is  perpendicular  to  the  other  also. 


N 


Given:  Plane  MN  J_  to  line  AB,  and  AB  II  to  CP. 

To  Prove :  CP  -L  to  plane  MN. 

Proof:  AB  and  CP  determine  a  plane.  (478.) 

Pass  this  plane  BC,  intersecting  plane  MN  in  line  BP. 

Draw  BX  J_  to  BP,  in  plane  MN. 

AB  is  J-  to  BX  (473). 

.  •.  BX  is  -L  to  plane  BC  (485). 

But  AB  is  J.  to  BP  (473). 

.-.  BP  is  J.  to  CP  (64). 

That  is,  plane  BC  is  J_  to  BX,  and  CP,  in  plane  BC,  is  _L  to 

the  intersection  BP. 

.-.  CP  is  J_  to  plane  MN        (492).     Q.E.D. 


274 


BOOK   VI.     SOLID   GEOMETRY 


PROPOSITION  XII.     THEOREM 
494.   Two  lines  perpendicular  to  the  same  plane  are  parallel. 


M, 


B 


'N 

Given :  Lines  AB  and  CD  _L  to  plane  MN. 
To  Prove :  AB\\io  CD. 

Proof :  Through  D,  the  foot  of  CD,  draw  DX  II  to  AB. 
Then  DX  is  J_  to  plane  MN  (493). 

But  CD  is  -L  to  plane  MN  at  D.  (Hyp.) 

.  •.  DX  and  DC  coincide  (490). 

That  is,  AB  is  II  to  CD.  Q.E.D. 

PROPOSITION  XIII.     THEOREM 

495.   Two  straight  lines  that  are  parallel  to  a  third  straight 
line  are  parallel  to  each  other. 


M, 

A 

E 

1 

\ 

j 

j 

s 

\ 

i 

! 

B 

D 

Given:  Lines  CD  and  .Breach  II  to  AB. 
To  Prove :  CD  II  to  EF. 

Proof :  Suppose  plane  MN  is  .passed  _L  to  AB. 
.-.  MN  is  _L  to  CD  and  to  EF 
.-.  CD  is  II  to  EF  (^94), 


(493). 

Q.E.I). 


LINES   AND   PLANES  275 

PROPOSITION  XIV.     THEOREM 

496.   A  line  perpendicular  to  one  of  two  parallel  planes  is 
perpendicular  to  the  other  also. 


Given :  Plane  MN  II  to  plane  ES ;  AP  -L  to  plane  RS. 
To  Prove :  AP  JL  to  plane  MN. 

Proof:  Through  AP  pass  any  two  planes,  AB  and  AC, 
intersecting  MN  in  AD  and  AE,  and  intersecting  RS  in  PB 
and  PC,  respectively. 

AD  is  II  to  PB,  and  AE  is  II  to  PC  (484). 

AP  is  J.  to  PB  and  PC  (473). 

.  •.  AP  is  _L  to  AD  and  AE  (64). 

.-.  AP  is  _L  to  plane  MN  (485).     Q.E.D. 

497.  COROLLARY.  If  two  planes  are  each  parallel  to  a  third 
plane,  they  are  parallel  to  each  other. 

Proof :  Draw  a  line  _L  to  the  third  plane. 
This  line  is  J_  to  each  of  the  other  planes  (496). 

.-.  The  two  planes  are  II  (489). 


Ex.  1.     Can  a  line  be  perpendicular  to  both  of  two  planes  if  they  are 
not  parallel  ?     Prove. 

Ex.  2.     Are  three   lines  that  are  perpendicular  to  the  same  plane 
necessarily  parallel  ? 

BOBBINS1  S   NEW    SOLID   GEOM. — 3 


276  BOOK   VI.     SOLID   GEOMETRY 

PROPOSITION  XV.     THEOREM 

498.   If  two  intersecting  lines  are  each  parallel  to  a  plane, 
the  plane  of  these  lines  is  parallel  to  the  given  plane. 


Given:  Intersecting  lines  AB  and  AC  in  plane  MN;  each 
line  II  to  plane  PQ. 

To  Prove :  Plane  MN  II  to  plane  PQ. 

Proof:  Draw  AE  -L  to  MN  at  A,  meeting  PQ  at  R. 
Through  AE  and  AB  pass  plane  AS,  and  through  AB  and 
AC  pass   plane   AT,  intersecting   plane  PQ   in    RS   and   ET> 
respectively. 

Now  AB  is  II  to  ES,  and  AC  is  II  to  ET  (483). 

AE  is  J_  to  AB  and  AC  (473). 

.-.  AE  is  _L  to  ES  and  ET  (64). 

Hence  AE  is  _L  to  plane  PQ  (485). 

.-.  plane  MN  is  II  to  plane  PQ  (489). 

Q.E.D. 

Ex.  1.  Can  one  line  be  perpendicular  to  two  other  lines  that  do  not 
intersect?  How? 

Ex.  2.  Can  one  line  be  perpendicular  to  two  other  lines  that  do  in- 
tersect? How? 

Ex.  3.  Could  484  be  quoted  correctly  as  the  reason  that  AB  is  par- 
allel to  E S  in  Proposition  XV  ? 

Ex.  4.  Could  483  be  quoted  correctly  as  the  reason  that  AD  is  par- 
allel to  BP  in  Proposition  XIV? 

Ex.  6.  Prove  Proposition  XV  by  drawing  AR  perpendicular  to  PQ 
from  A. 


LINES   AND   PLANES  277 

PROPOSITION  XVI.     THEOREM 

499.  If  two  angles,  not  in  the  same  plane,  have  their  sides 
parallel  each  to  each,  and  extending  in  the  same  directions  from 
their  vertices,  the  angles  are  equal  and  the  planes  are  parallel. 

N 


Given:  /.BAG  in  plane  MN  and  Z  EDF  in  plane  PQ;  AB  II 
to  DE-,  AC  II  to  DF,  and  extending  in  the  same  directions. 
To  Prove :  I.     Z  BAG  =  Z  EDF. 
II.     Plane  MN  II  to  plane  PQ. 

Proof:  I.     Take  DE  and  AB  equal,  and  DF  and  AC  equal. 
Draw  AD,  BE,  CF,  BC,  EF. 
The  figure  ABED  is  a  O  (129). 

.-.  AD  =  BE  (124). 

Also  ACFD  is  a  O  (?). 

AD  =  CF  (?). 

.-.  BE=CF  (?). 

Again,  AD  is  II  to  BE  and  AD  is  II  to  CF  (120). 

.-.  BE  is  II  to  CF  (495). 

.-.  BCFEis  a  O  (129). 

Now  in  &ABC  and  DEF,  AB  =  DE',  AC  =  DF         (Const.). 

Also  BC  =  EF  (124). 

(78). 

(27). 
Q.E.D. 

II.     AB  is  II  to  plane  PQ  and  AC  is  II  to  plane  PQ       (482). 
.-.  plane  MN  is  II  to  plane  PQ  (498). 

Q.E.D. 


278  BOOK   VI.     SOLID   GEOMETRY 

PROPOSITION  XVII.     THEOREM 

500.   If  three  parallel  planes  intersect  two  straight  lines, 
the  corresponding  intercepts  are  proportional. 


5 


Given:  Parallel  planes,  L3/,  NP,  QR,  intersecting  line  AB 
at  A,  E,  J5,  and  CD  at  C,  F,  D,  respectively. 

To  Prove  :  AE  :  EB  =  CF  :  FD. 

Proof  :  Draw  BC,  meeting  plane  NP  at  G. 

Through  AB  and  BC  pass  a  plane  cutting  LM  in  AC  and 
NP  in  EG. 

Through  BC  and  CD  pass  a  plane  cutting  NP  in  GF  and  QR 
in  BD. 

Now  EG  is  II  to  ^1(7  and  GF  is  II  to  £D  (484). 


.EB        GJ5 

Consequently          AE  :  En  =  CF  :  FD  (Ax.  1). 

Q.E.D. 

Also  AE  +  EB  :  AE  =  CF  +  FD 

Or  AE  +  J££  :  EB  =  C.P  +  FD 

.'.  AB  :  AE=  CD  :  CF  } 
Or  AB:EK=CD:FD 


CD       CF       FD 


... 

:  CF  1 
:  FD  J 


Ex.  1.  If  any  number  of  lines  which  meet  at  a  point  are  cut  by 
two  parallel  planes,  the  corresponding  intercepts  are  proportional. 

Ex.  2.  In  the  above  diagram,  why  are  not  AC  and  BD  parallel? 
Under  what  condition  would  EOF  be  a  straight  line  ? 


LINES   AND   PLANES  279 


PROPOSITION  XVIII.     THEOREM 

501.   The  projection  of  a  straight  line  upon  a  plane  is  a  straight 
line.* 


T 


R  J 


—N 

Given :  Line  ^17?  and  plane  MN. 
To  Prove :  The  projection  of  AB  on  MN  is  a  straight  line. 

Proof :  Draw  PJ  _L  to  plane  MN  from  any  point  P,  in  AB. 

AB  and  PJ  determine  a  plane.  (479). 

Plane  AD  cuts  plane  MN  in  a  straight  line  CD.  (481). 

Now  in  plane  AD,  draw  XE  II  to  PJ  from  X,  any  other 
point  in  AB. 

XE  is  -L  to  plane  MN.  (493). 

Now  E  is  the  projection  of  X.  (469). 

.*.  CD  is  the  projection  of  AB.  (469). 

That  is,  the  projection  of  AB  upon  the  plane  MN  is  a 
straight  line.  Q.E.D. 

Ex.  1.  What  is  the  length  of  the  projection  of  a  5  ft.  rod,  inclined  at 
an  angle  of  45°  ? 

Ex.  2.  A  ladder  26  ft.  long  leans  against  the  wall  of  a  house,  at  a 
point  10  ft.  from  the  ground.  What  is  the  length  of  the  projection  of 
the  ladder  on  the  ground? 

502.  COROLLARY.  A  straight  line  and  its  projection  upon  a 
plane  are  in  the  same  plane. 

*  Except  only  if  the  given  line  is  a  normal  to  the  given  plane. 


280 


BOOK   VI.     SOLID   GEOMETRY 


PROPOSITION  XIX.     THEOREM 

503.   A  line  not  parallel  to  a  plane  is  longer  than  its  projection 
upon  the  plane. 

N 


Given:   A  plane  and  line  LN  not  II  to  the  plane,  and  DE 
the  projection  of  LN  upon  the  plane. 

To  Prove :  LN  >  DE. 

Proof :  Draw  LD  and  NE. 
Draw  LX  -L  to  NE  from  z,  in  the  plane  LE. 
LD  and  NE  are  JL  to  the  plane.     (Def.  of  projection,  469). 
LXED  is  a  rectangle.  (157). 

Now  LN  >  LX  (87). 

But  LX=  DE  (124). 

.-.  LN  >  DE  (Ax.  6).       Q.E.D. 


Ex.     If  a  line  is  parallel  to  a  plane,  all  points  of  the  line  are  equally 
distant  from  the  plane. 

PROPOSITION  XX.     THEOREM 

504.   Of  all  lines  that  can  be  drawn  to  a  plane  from  a  point : 
I.  The  perpendicular  is  the  shortest. 
II.  Oblique  lines  having  equal  projections  are  equal. 
HI.   Equal  oblique  lines  have  equal  projections. 
IV.   Oblique  lines  having  unequal  projections  are  unequal, 
and  the  line  having  the  greater  projection  is  the  longer. 

V.   Unequal  oblique  lines  have  unequal  projections,  and 
the  longer  line  has  the  greater  projection. 


LINES   AND   PLANES 
P 


281 


'N 


I.    Given :  Plane  MN ;  point  P  ;  PR  _L  to  MN ;  any  other 
line  from  P  to  plane  MN,  as  PA. 
To  Prove :  PR  <  PA. 
Proof:  Draw  AE. 

Now  PB  is  J.  to  AE  (473). 

And  PA  is  not  _L  to  ^412  (54). 

.-.  P£  <  P^i  (87).     Q.E.D. 

II.    Given:  Oblique  lines  PA  and  PB  whose  projections, 
AE  and  BE,  are  equal. 
To  Prove :  PA  =  PB. 
Proof:  The  right  A  PAR  and  PEB  are  ^  (?). 

III.  Given :  Equal  oblique  lines  PA  and  PB. 

To  Prove :  Their  projections,  AE  and  BR,  are  equal. 
Proof :  The  right  A  PAR  and  PEB  are  ^  (?). 

IV.  Given :  Oblique  lines  PC  and  PA  ;  proj.  EC  >  proj.  l?^. 
To  Prove :  PC  >  PA. 

Proof :  In  A  PRC,  take  on  RC,  EX  =  EA,  and  draw  PX. 
Now  PC  >  PX  (88,  III). 

But  PA  =  PX  (504,  II). 

.-.  PC  >  PA  (Ax.  6).     Q.E.D. 

V.    Given :  Unequal  oblique  lines,  PC  >  PA. 
To  Prove :  Projection  EC  >  projection  EA. 
Proof:  By  method  of  exclusion  (See  90). 


2*2  BOOK  VI.     SOLID   GEOMETRY 

PROPOSITION  XX F.     THEOREM 

505.  The  acute  angle  that  a  line  makes  with  its  own  projec- 
tion upon  a  plane  is  the  least  angle  that  the  line  makes  with 
any  line  of  the  plane. 

A 


Given :  AB,  any  line  meeting  plane  MN  at  B  ;  BP,  its  projec- 
tion upon  MN ;   BD,  any  other  line  in  JfJV,  through  B. 

To  Prove :  Z  ABP  <  Z.  ABD. 

Proof :  On  BD  take  BX  =  BP  and  draw  AX. 
In  A  APB  and  ABX, 

AB  =  AB  (?). 

BP—BX  (Const.). 

But                                   AP  <  AX  (504,  I). 

.-.  Z-ABP  <  Z.ABD  (92). 

Q.E.D. 


Ex.  1.     If  PR  is  12  in.,  and  AP  13  in.,  find  the  length  of  AR,  in  504. 

Ex.  2.     Does  the  longer  of  two  lines  always  have  the  longer  projection 
on  the  same  plane?     Could  they  have  equal  projections?     Illustrate. 

Ex.  3.     With  what  line  in  a  plane  does  a  line  oblique  to  that  plane 
make  the  greatest  angle  ? 

Ex.  4.     With  what  line  in  a  plane  does  a  line  oblique  to  that  plane 
make  right  angles  ? 

Ex.  5.     How  do  you  construct  the  projection  of  a  curved  line  upon  a 
plane?     When  is  this  a  straight  line? 


LINES   AND  PLANES  283 

PROPOSITION  XXII.     THEOREM 

606.   Through  a  given  point  one  plane  can  be  passed  parallel 
to  any  two  given  non-parallel  lines  in  space,  and  only  one. 

B 
A 


M 


Given:  Point  P;  two  lines,  AB  and  CD. 

To  Prove :  Through  P  one  plane  can  be  passed  II  to  AB  and 
CD,  and  only  one. 

Proof:  I.  Through  P  draw  a  line  II  to  AB  and  another 
II  to  CD. 

Pass  a  plane  MN,  containing  these  lines. 

MN  is  II  to  both  AB  and  CD  (482). 

II.    Only  one  line  can  be  drawn  through  P  II  to  AB,  and 

only  one  II  to  CD  (Ax.  13). 

.-.  there  is  only  one  plane  (479). 

Q.E.D. 

507.  COROLLARY.  If  two  lines  are  not  in  the  same  plane,  one 
plane  and  only  one  can  be  passed  through  one  of  these  lines 
parallel  to  the  other. 

[Through  a  point  in  one  line  draw  a  line  ||  to  the  other  line,  etc.] 

Ex.  1.  Is  Proposition  XXII  true  if  the  given  lines  intersect  ?  Is  it 
true  if  they  are  parallel  ? 

Ex.  2.  Explain,  so  that  a  blind  boy  could  understand,  how  to  pass  a 
plane  through  a  given  point  and  parallel  to  two  pencils  he  may  hold  in 
his  outstretched  hands. 

Ex.  3.  Can  two  lines  be  parallel  to  a  plane  and  not  be  parallel  to 
each  other?  Illustrate,  by  means  of  pencils  and  the  ceiling. 


284 


BOOK   VI.     SOLID   GEOMETRY 


PROPOSITION  XXIII.     THEOREM 

508.  Through  a  given  point  one  plane  can  be  passed  parallel 
to  a  given  plane,  and  only  one. 


/ 


Given:  (?).     To  Prove:  (?). 

Proof :  I.     Suppose  PR  is  drawn  J_  to  plane  AB  ;  and  plane 
XY  is  passed  J_  to  PR  at  P. 

Then                            XY  is  II  to  AB  (489). 

II.     Only  one  line  -L  to  AB  can  be  drawn  from  P  (491). 

Only  one  plane  _L  to  PR  can  be  passed  at  P  (^87). 

.•.  only  one  plane  can  contain  P  and  be  II  to  AB.  Q.E.D. 

PROPOSITION  XXIV.     THEOREM 

609.  Parallel  lines  included  between  parallel  planes  are 
equal. 


Given:  (?).     To  Prove:  (?). 

Proof:       The  plane  determined  by  AB  and  CD  intersects 

RS  and  PQ  in  lines  AC  and  BD,  which  are  II  (484). 

.-.  ABDC  is  a  O  (Def.). 

Hence  AB  =  CD  (124).     Q.E.D. 


LINES   AND  PLANES 


285 


PROPOSITION  XXV.     THEOREM 

610.  The  plane  perpendicular  to  a  line  at  its  midpoint  is  the 
locus  of  points  in  space  equally  distant  from  the  extremities  of 
the  line. 


A^ 


M 


,'P' 


B 


Given :  Plane  RS  _L  to  AB  at  its  midpoint,  M. 

To  Prove :  Plane  RS  is  the  locus  of  points  in  space  equally 
distant  from  A  and  B. 

Proof:  (1)  Take  P,  any  point  in  RS. 

Draw  PM,  PA,  PB. 

Now  PM  is  J.  to  AB  (473). 

.-.  PA  =  PB  (80). 

That  is,  any  point  in  RS  is  equally  distant  from  A  and  B. 

(2)  Take  P',  any  point  outside  of  RS.     Draw  PfM. 

Now  P'M  is  not  -L  to  AB  (486). 

.•.  P',  any  point  outside  of  plane  RS,  is  not  equally  distant 
from  A  and  B  (81). 

.-.  Plane  RS  is  the  locus  of  points  in  space,  equally  dis- 
tant from  A  and  B  (246).  Q.E.D. 


Ex.  1.  How  can  you  find  a  line  in  a  plane  such  that  each  of  its 
points  is  equally  distant  from  two  given  points? 

Ex.  2.  How  can  you  find  that  point  in  a  given  line  which  is  equally 
distant  from  two  given  points  ? 

Ex.  3.  There  are  two  definite  lines  in  space.  It  is  desired  to  find  all 
points  that  are  equally  distant  from  the  ends  of  one  line  and  at  the  same 
time  equally  distant  from  the  ends  of  the  other.  How  can  this  be  done? 


286 


BOOK  VI.     SOLID   GEOMETRY 


PROPOSITION  XXVI.     THEOREM 

511.  The  locus  of  points  in  space  equally  distant  from  all 
the  points  in  the  circumference  of  a  circle  is  the  line  perpen- 
dicular to  the  plane  of  the  circle  at  its  center. 

A 


N 


M 

Given:  (?).     To  Prove:  (?). 

Proof:  I.     Any  point  in  AC  is  equally  distant  from  all  the 

points  in  the  circumference  of  the  circle  (504,  II). 

II.     Any  point   equally  distant   from   all   points    of   the 

circumference  of  the  circle  is  in  AC  (504,  III). 

.-.  AC  is  the  required  locus  (246).     Q.E.D. 


Ex.  1.  What  is  the  locus  of  points  equally  distant  from  two  given 
points  ? 

Ex.  2.  What  is  the  locus  of  points  equally  distant  from  three  given 
points  ? 

Ex.  3.  Draw  a  triangle  on  the  blackboard  and  a  definite  straight  line 
on  the  floor.  Tell  how  to  find  the  one  point  which  is  both  equally  dis- 
tant from  the  vertices  of  the  triangle  and  from  the  ends  of  the  line.  Is 
there  always  one  point? 

512.  The  distance  from  a  point  to  a  plane  is  the  length  of 
the  perpendicular  from  the  point  to  the  plane. 

Thus,  the  word  "distance,"  referring  to  the  shortest  line  from  a  point 
to  a  plane,  implies  the  perpendicular.  ^ 

The  inclination  of  a  line  to  a  plane  is  the  angle  between 
the  line  and  its  projection  upon  the  plane. 


ORIGINAL   EXERCISES  287 


ORIGINAL    EXERCISES 

1.  Through  one  straight  line  a  plane  can  be  passed  parallel  to  any 
other  straight  line  in  space,  and  only  one. 

Through  a  point  of  the  first  line  draw  a  line  II  to  the  second. 

2.  Two  parallel  planes  are  everywhere  equally  distant. 

3.  If  a  line  and  a  plane  are  parallel,  another  line  parallel  to  the  given 
line  and  through  any  point  in  the  given  plane  lies  wholly  in  the  given 
plane. 

Through  the  given  line  and  the  point  P  pass  a  plane  cutting  the 
given  plane  in  PX.  [Use  483.] 

4.  A  straight  line  parallel  to  the  intersection  of  two  planes,  but  in 
neither,  is  parallel  to  both  planes. 

5.  If  two  straight  lines  are  parallel  and  two  intersecting  planes  are 
passed,  each  containing  one  of  the  lines,  the  intersection  of  these  planes 
is  parallel  to  each  of  the  given  lines. 

6.  If  three  straight  lines  through  a  point  meet  the  same  straight  line, 
these  four  lines  all  lie  in  the  same  plane. 

7.  If  a  straight  line  meets  two  parallel  planes,  its  inclinations  to  the 
planes  are  equal. 

8.  Two  parallel  planes  can  be  passed,  each  containing  one  of  two 
given  lines  in  space.     Is  this  ever  impossible? 

9.  If  each  of  three  straight  lines  intersects  the  other  two,  the  three 
lines  all  lie  in  a  plane. 

10.  The  projections  of   two  parallel    lines  on  a 
plane  are  parallel. 

Proof:  AB  is  II  to  CD  (?).     AE  is  II  to  CG  (?).  . 

.-.  planes  AF  and  CH  are  II  (?)  ;  etc.  N 

11.  If  two  lines  in  space  are  equal  and  parallel,  their  projections  on  a 
plane  are  equal  and  parallel. 

12.  If  a  plane  is  parallel  to  one  of  two  parallel  lines,  it  is  parallel  to 
the  other. 

13.  If  a  straight  line  and  a  plane  are  perpendicular  to  the  same  straight 
line,  they  are  parallel. 

14i  Equal  oblique  lines  drawn  to  a  plane  from  one  point  have  equal 
inclinations  with  the  plane. 

15.  If  a  line  and  a  plane  are  both  parallel  to  the  same  line,  they  are 
parallel  to  each  other. 


Mi — j — JQ 
/sUp 


288 


BOOK   VI.     SOLID   GEOMETRY 


16.   Four  points  in  space,  A,  B,  C,  D,  are  joined,  and  these  four  lines 
are  bisected.     Prove  that  the  four  lines  joining  (in 
order)  the  four  midpoints  of   the  first  lines  form  a 
parallelogram. 

Proof:  Pass  plane  DP  through  points  A,  D,  B, 
and  plane  DX  through  points  B,  C,  D,  —  these 
planes  intersecting  in  BD.  ST  is  ||  to  BD  and  = 
?):  etc. 


17.  If   a  plane  is  passed  containing  a  diagonal  of  a  parallelogram 
and  perpendiculars  are  drawn  to  the  plane  from  the 

other  vertices  of  the  parallelogram,  they  are  equal. 
To  Prove:     AE  =  CF.     Proof:    Draw  diagonal 
AC.     Draw  EO  and  OF  in  plane  MN.     EO,  OF, 
and  EOF  are  projections;  etc. 

18.  If    from   the  foot   of   a  perpendicular  to   a 
plane,  a  line  is  drawn  at  right  angles  to  any  line 
in  the  plane,  the  line  connecting  this  point  of  inter- 
section with  any  point  in  the  perpendicular  is  per- 
pendicular to  the  line  in  the  plane. 

Given:  AE  _L  to  plane  RS ;  EC  L  to  DE  in  the 
plane;  PC  drawn  from  C  to  P,  in  AE. 

To  Prove :   PC  is  J_  to  DE. 

Proof :  Take  CD  =  CE,  draw  PD,  PE,  BD,  BE.    EC  is  _L  to  DE  at 

its  midpoint  (?).     .-.  ED  =  BE  (?).    PD  =  PE  (?)  (504,  II). 

.-.  PC  is  ±  to  DE  (?)  (83). 

19.  A  line  PE  is  perpendicular  to  a  plane  at  B,  and  a  line  is  drawn 
from  B  meeting  any  line  DE,  of  the  plane,  at  C.     If  PC  is  perpendicular 
to  DE,  BCis  perpendicular  to  DE. 

20.  Are    two    planes    that   are  parallel  to  the   same  straight    line 
necessarily  parallel? 

21.  -If  each  of  two  parallel  lines  is  parallel  to  a  plane,  is  the  plane  of 
these  lines  also  parallel  to  the  given  plane? 

22.  Is  a  three-legged  stool  always  stable  on  the  floor?     Why?    Is  a 
four-legged  chair  always  stable?     Why? 

23.  What  is  the  locus  in  space  of  points  equally  distant  from  two 
parallel  planes?  from  two  parallel  lines? 

24.  What  is  the  locus  of  points  in  space  at  a  given  distance  from  a 
given  plane? 


ORIGINAL   EXERCISES  289 

26.   What  is  the  locus  of  points  in  a  plane  at  a  given  distance  from 
an  external  point? 

26.  What  is  the  locus  of  points  in  space  equally  distant   from  two 
points  and  equally  distant  from  two  parallel  planes? 

27.  What  is  the  locus  of  points  in  space  equally  distant  from  the  ver- 
tices of  a  triangle  ? 

28.  What  is  the  locus  of  all  straight  lines  perpendicular  to  a  given 
straight  line  at  a  given  point  ? 

29.  What  is  the  locus  of  all  lines  parallel  to  a  given  plane  and  drawn 
through  a  given  point? 

30.  If  the  points  in  a  line  satisfy  one  condition  and  the  points  in  a 
plane  satisfy  another  condition,  what  will  be  true  of  their  intersection? 
What  will  be  true  if  they  do  not  intersect? 

31.  If  the  points  in  one  plane  satisfy  one  condition  and  the  points  in 
another  plane  satisfy  another  condition,  what  is  true  of  their  intersection  V 
What  is  true  if  the  planes  are  parallel? 

32.  Construct  a  plane  perpendicular  to  a  given  line  at  a  given  point 
in  the  line. 

33.  Construct  a  plane  perpendicular  to  a  given  line  through  a  given 
external  point. 

34.  Construct  a  line  perpendicular  to  a  given  plane  through  a  given 
point  in  the  plane ;  through  a  given  external  point. 

36.    Construct  a  plane  parallel  to  a  given  plane  through  a  given  point. 

36.  Construct  a  number  of  equal  oblique  lines  to  a  plane  from  a  given 
external  point. 

37.  Construct  a  line  through  a  given  point  parallel  to  a  given  plane. 

38.  Construct  through  a  given  point  a  line  parallel  to  each  of  two 
given  intersecting  planes. 

39.  Construct   a  plane  containing  one   given   line  and  parallel    to 
another. 

40.  Construct  a  plane  through  a  given  point  parallel  to  any  two  given 
lines  in  space. 

41.  Construct  a  line  through  a  given  point  in  space  which  intersects 
two  given  lines  not  in  the  same  plane. 

When  is  there  no  such  line  ?    Is  there  ever  more  than  one  ? 

42.  Find  a  point  in  a  plane  such  that  the  sum  of  the  two  lines  join- 
ing it  to  two  fixed  points  on  one  side  of  the  plane  is  the  least  possible. 


290  BOOK   VI.     SOLID   GEOMETRY 

Construction:    Draw    A(.'   -L   to    plane   ^fN    and  yB 

prolong    it    to    A",    making-    CX  =  AC.       Draw    BX,  A  / 

meeting   plane   MN  at    P.     Draw    A  P.     Take   any  .  \       / .N 

other  point  R  in  plane  MX.  I  \  \/      R\ 

Statement :  .4P  +  PJ3  <  .4  A'  +  725.     Etc.  /          P         \ 

43.  Find  a  point  in  a  given  plane  equally  distant          yj/ 
from  three  given  points.     Is  this  ever  impossible? 

44.  Find  the  one  point  equally  distant  from  four 
given  points  not  in  the  same  plane. 

Construction:  Pass  plane  CM,  containing  points  A, 
B,  C,  and  plane  CN,  containing  A,  Z>,  C.  Find  0,  the 
center  of  the  O  containing  A,  B,  C.  Find  P,  simi- 
larly. Draw  the  locus  of  points  equally  distant  from 
A,  By  C.  (Consult  511.)  Draw  the  locus  of  points 
equally  distant  from  A,  D,  C.  The  plane  _L  to  AC  at 
its  midpoint  contains  both  these  loci.  (Explain.)  Hence  OX  and  PA' 
intersect  (?).  .•.  X  is  the  required  point. 


DIHEDRAL   ANGLES 

513.  A    dihedral    angle     is     the  E^^^B^MK     -    1 A 
amount  of  divergence  of  two  inter- 
secting planes.     The   edge   of    the  ft 
dihedral  angle  is  the  line  of  inter-  QJ^  _Jp 
section   of    the   planes.     The  faces 

of  the  dihedral  angle  are  the  planes. 

The  intersecting  planes  AG  and  ED  form  the  dihedral  angle  whose 
edge  is  EG,  which  is  named  A-GE-D;  or,  when  there  is  only  one 
dihedral  angle  at  the  edge,  "the  angle  EG." 

514.  Adjacent  dihe- 
dral   angles    are    two 
dihedral    angles    that 
have   the    same    edge 
and    a    common    face 
between  them. 


DIHEDKAL    ANGLES  291 

Vertical  dihedral  angles  are  two  dihedral  angles  that  have 
the  same  edge,  the  faces  of  one  being  the  extensions  of  the 
faces  of  the  other. 

515.  The  plane  angle  of  a  dihedral  angle  is  the  angle 
formed  by  two  straight  lines,  one  in  each  face,  and  perpen- 
dicular to  the  edge  at  the  same  point. 

If  PM  is  in  plane  AG  and  perpen- 
dicular to  EG,  and  PN  is  in  plane  ED 
and  perpendicular  to  EG  at  P,  the 
angle  MPN  is  the  plane  angle  of  the 
dihedral  angle  EG. 


516.  If   one  plane   meets   another,  making   the   adjacent 
dihedral    angles    equal,    these    angles    are    right    dihedral 
angles. 

One  plane  is  perpendicular  to  another  plane  if  the 
dihedral  angle  formed  by  the  two  planes  is  a  right  dihedral 
angle. 

517.  Two  dihedral  angles  are  equal  if  they  can  be  made 
to  coincide. 

A  dihedral  angle  is  acute,  right,  or  obtuse  according  as  its 
plane  angle  is  acute,  right,  or  obtuse. 

Dihedral  angles  are  complementary  or  supplementary,  cor- 
responding, alternate-interior,  etc.,  according  as  their  plane 
angles  are  complementary  or  supplementary,  corresponding, 
alternate-interior,  etc. 

NOTE.  An  open  book  often  assists  a  student  to  a  clear  apprehension 
of  the  magnitude  of  dihedral  angles.  Thus  he  can  see  the  angle  increase 
during  the  act  of  opening  the  book,  and  observe  the  acute,  right,  and 
obtuse  dihedrals.  With  the  aid  of  two  books  he  can  understand  better, 
perhaps,  the  meaning  of  complementary  dihedrals,  supplementary  di- 
hedrals, corresponding  dihedrals,  alternate-interior  dihedrals,  etc. 

ROBBINS'S    NEW    SOLID    GEOM.  —  4 


292 


BOOK   VI.     SOLID   GEOMETRY 


PROPOSITION  XXVII.    THEOREM 

518.  The  plane  angles  of  a  dihedral  angle  are  all  equal. 

A 

E 
R 

Given :  Z  EFG,   the  plane  Z  of   dihedral  Z  .BC7,  at  .F,  and 
Z  RST,  the  plane  Z  at  <S. 
To  Prove :  Z  ^FG  =  Z  RST. 

Proof:  EFis  II  to  RS  (62). 

And  FG  is  II  to  ST  (?). 

.*.  Z  im?  =  Z  .RST  (499).    Q.E.D. 

519.  COROLLARY.    The  plane  of  the  plane  angle  of  a  dihe- 
dral angle  is  perpendicular  to  the  edge.  (485). 

PROPOSITION  XXVIII.    THEOREM 

520.  Two  dihedral  angles  are  equal  if  their  plane  angles  are 

e^'  i  i  \A 


E    B 


\l, 


JE' 


Given:  Dihedral  A  CB  and  c' B'  whose  plane  A EBD  and 
ErBfDf  are  equal. 

To  Prove :  Dih.  Z  CB  =  dih.  Z  (/£'. 

Proof :  CJ5  is  ±  to  plane  EBD. 

And  C/jB'  is  J_  to  plane  ^'B'D'  (519). 


DIHEDRAL   ANGLES  293 

Apply  dih.  Z  C'B'  to  dih.  Z  CB  so  that  the  plane  Z  E'B'D1 
coincides  with  its  equal  Z  EBD. 

Now  C'B'  coincides  with  CB  (490). 

.-.  plane  C'D'  coincides  with  plane  CD  and  plane  CrEf  coin- 
cides with  plane  CE  (479). 
.  •.  dih.  Z  CB  =  dih.  Z  C'B'  (517). 

Q.E.D. 

PROPOSITION   XXIX.     THEOREM 

521.  If  two  dihedral  angles  are  equal,  their  plane  angles  are 
equal.     [Converse.] 

Proof:  Superpose  dih.  Z.C'B'    upon   its   equal  dih.  Z.  CB' 
making  B'  fall  on  jB,  and  edge  BfCf  on  BC. 
Then  face  C'D'  coincides  with  face  CD,  etc. 

522.  COROLLARY.     Two  vertical  dihedral   angles   are  equal. 

(See  520.) 

523.  COROLLARY.     The  plane  angle  of  a  right  dihedral  angle 
is  a  right  angle ;  and  if  the  plane  angle  of  a  dihedral  angle  is  a 
right  angle,  the  dihedral  angle  is  right.  (See  516.) 

PROPOSITION  XXX.     THEOREM 

524.  Two  dihedral  angles  have  the  same  ratio  as  their  plane 
angles. 


A   C 


qG; 

B, 


A 


Given:  Dihedral  A  A-BC-D  and  A'-B'C'-D',  having  plane 
A  ACE  and  A'C'E',  respectively. 
To  Prove : 
Dih.  Z  A-BC-D  :    dih.  Z  A'-B'c'-D'  =  Z  ACE :   Z 


294 


BOOK   VI.     SOLID   GEOMETRY 


Proof:  I.    If  the  plane  angles  are  commensurable. 
There  exists  a  common  unit  of  measure  of  the  plane  A  ACE 
and  A'C'E'  (224). 


B 


E| 


JA   C| 

B'l 


Suppose  this  unit  when  applied  to  these  angles  is  contained 
3  times  in  Z  ACE  and  4  times  in  Z  A'C'E'. 

ZACE  _3  (,      ox 

•    •    ; — ; — ; — "7  I  /\X.    O). 

Z  ACE        4 

Pass  planes  through  the  edges  and  the  several  lines  of  divi- 
sion of  the  angles. 

Dih.  Z  A-BC-D  is  divided  into  3  parts  ;  dih.  Z  A'-B'C'-D'  is 
divided  into  4  parts  ;  all  of  these  seven  parts  are  equal  (520). 

dih.  Z  A-BC-D        3  ,  .       ON 

•"•  ~rn ; — ~> — '• — ~< — T  =  T  ^AX.  o). 

dih.  /.A'-B'C'-D'     4 

.  •.  dih.  Z  A-BC-D  :  dih.   Z  A'-B'c'-D1  =  Z  ACE :    Z  A'C'E' 

(Ax.  1). 
II.    If  the  plane  angles  are  incommensurable. 

C 


B 


There  does  not  exist  a  common  unit.  Suppose  £  ACE  to  be  divided 
into  equal  parts  (any  number  of  them). 

Apply  one  of  these  as  a  unit  of  measure  to  Z.  A'C'E'.  There  is  a 
remainder,  XC'E',  left  over  (because  the  A  are  incommensurable). 

Pass  a  plane  C'Y,  determined  by  B'C'  and  C'X. 


DIHEDRAL   ANGLES 


295 


Now 


dih.  /.A-BC-D        /.ACE 


di\}./.A'-B'C'-Y     /.A'C'X 


(Commensurable 


Indefinitely  increase  the  number  of  subdivisions  of  /.  A  CE. 

Then  each  part,  that  is,  our  unit  or  divisor,  is  indefinitely  decreased. 
Hence  XC'E',  the  remainder,  is  indefinitely  decreased. 

That  is,  /.  XC'E'  approaches  zero  as  a  limit. 

And  dih.  /.X-B'C'-D1  approaches  zero  as  a  limit. 

.-.  /.A'C'X  approaches  A'C'E'  as  a  limit;  and  dih.  /.A'-B'C'-Y 
approaches  dih.  /.A'-B'C'-D'  as  a  limit. 


dih.  /.  A'-B'C'-Y  ~  dih.  Z  A'-B'C'-D' 

approaches- — r^r4r.as  a  limit, 
dih.  /.A-BC-D 


/.ACE 


dih.^A'-B'C'-D'     /.A'C'E' 


(229).     Q.E.D. 


PROPOSITION  XXXI.     THEOREM 

525.   If  a  straight  line  is  perpendicular  to  a  plane,  any  plane 
containing  this  line  is  perpendicular  to  the  given  plane. 


Given:  Line  AB  _L  to  plunc  MN ;  plane  PQ  containing  AB 
and  intersecting  plane  MN  in  RS. 

To  Prove :  Plane  PQ  is  _L  to  plane  MN. 

Proof :  In  plane  M N  draw  BC  -L  to  RS. 

Now                              AB  is  -L  to  RS.  (473). 

.-.  Z  ABC  is  the  plane  Z  of  dih.  Z  p-SR-N.  (515). 

But                             Z  ^BC  is  a  rt.  Z.  (473). 

.-.  PQ  is  J.  to  MN  (516).     Q.E.D. 


296  BOOK   VI.     SOLID   GEOMETRY 

526.  COROLLARY.     If  a  plane  is  perpendicular  to  the  edge  of 
a  dihedral  angle,  it  is  perpendicular  to  each  face.      (See  525.) 

PROPOSITION   XXXII.     THEOREM 

527.  If  one  plane  is  perpendicular  to  another,  any  line  in 
either  plane,  perpendicular  to  their  intersection,  is  perpendicular 
to  the  other  plane.     [Converse  of  525.] 


Given:  Plane  PQ  ±  to  plane  MN;  AB  in  plane  PQ  _L  to  tl.. 
intersection,  RS. 

To  Prove :  AB  _L  to  plane  MN. 
Proof :  In  plane  MN  draw  BC  J_  to  RS. 
Now  Z  ABC  is  the  plane  angle  of  the  dih.  Z  P-SR-N.     (515). 

.-.  Z  ABC  is  art.  Z  (523). 

.-.  AB  is  J_  to  BC  (16). 

But                              AB  is  _L  to  RS  (Hyp.)- 

.-.  AB  is  J_  to  plane  MN  (485). 

Q.E.D. 

Ex.  1.     In  the  figure  of  527  prove  BC  perpendicular  to  the  plane  PQ. 
Also  prove  RS  perpendicular  to  the  plane  ABC. 

Ex.  2.     In  the  figure  of  527  prove  plane  ABC  perpendicular  to  the 
planes  MN  and  PQ. 

Ex.  3.     Under  what  condition  will  a  line  in  one  face  of  a  dihedral 
angle  meet  a  line  in  the  other  face  ? 


DIHEDRAL   ANGLES 


297 


PROPOSITION  XXXIII.     THEOREM 

528.  If  one  plane  is  perpendicular  to  another,  a  line  drawn 
from  any  point  in  their  intersection  and  perpendicular  to  one 
plane,  lies  in  the  other. 


Given:  Plane   PQ  _L   to   plane   MN,    intersecting   in   R8-, 
AS  (left-hand)  J_  to  plane  MN  from  A,  in  RS. 

To  Prove :  AB  is  in  plane  PQ. 

Proof :  At  A  erect  in  plane  PQ,  AX  JL  to  RS. 
Then  AX  is  -L  to  plane  MN 

But  AB  is  -L  to  plane  MN  at  A 

.•.  AB  and  AX  coincide 
That  is,  AB  lies  in  plane  PQ. 

PROPOSITION  XXXIV.     THEOREM 


(527). 

(Hyp.)- 
(490). 

Q.E.D. 


529.  If  one  plane  is  perpendicular  to  another,  a  line  drawn 
from  any  point  in  one  plane,  and  perpendicular  to  the  other, 
lies  hi  the  first  plane. 

Given:  Plane  PQ  _L  to  plane  MN ;  AB  (right-hand)  _L  to 
plane  MN  from  A,  any  point  in  plane  PQ. 

To  Prove :  AB  lies  in  plane  PQ. 

Proof:  From  A  draw  in  plane  PQ,  AX  A.  to  RS. 

Then                      AX  is  -L  to  plane  MN  (527). 

.-.  AB  and  AX  coincide  (491). 

That  is,                   AB  lies  in  plane  PQ.  Q.E.D. 


298  BOOK  VI.     SOLID   GEOMETRY 

PROPOSITION  XXXV.     THEOREM 

530.   If  two  planes  are  perpendicular  to  a  third  plane,  their 
intersection  also  is  perpendicular  to  that  plane. 


Given :  Planes  LM  and  2VP,  each  _L  to  plane  ES. 
To  Prove :  The  intersection  AB  is  _L  to  plane  BS. 
Proof :  If,  at  A,  a  line  is  erected  _L  to  plane  RS,  it  will  lie 

in  plane  LM  (528). 

This  _L  will  lie  also  in  plane  NP  (?). 

.-.  this  -L  is  the  intersection  AB  (466). 

That  is,                     AB  is  _L  to  plane  RS.  Q.E.D. 

531.  COROLLARY.  If  a  plane  is  perpendicular  to  each  of  two 
intersecting  planes,  it  is  perpendicular  to  their  intersection. 
(The  same  truth  as  530.) 


Ex.  1.  If  each  of  three  planes  is  perpendicular  to  the  other  two, 
each  of  the  three  intersections  is  perpendicular  to  the  remaining  plane, 
and  perpendicular  to  the  other  two  intersections. 

[Use  figure  of  530.] 

Ex.  2.  If  two  parallel  planes  are  each  perpendicular  to  a  third  plane, 
their  intersections  with  that  plane  are  parallel. 

Ex.  3.  Is  Proposition  XXXV  true  in  the  case  of  the  intersecting 
walls  of  a  building? 

Ex.  4.  What  is  the  "  exterior  point "  that,  together  with  the  plumb- 
bob,  determines  the  vertical  plane  for  the  mason  ? 

Ex.  5.  If  the  builder  keeps  the  edge  of  a  building  perpendicular  to 
the  plane  of  a  level  street,  the  two  intersecting  walls  will  also  be  perpen- 
dicular to  the  street.  Why  ? 


DIHEDRAL   ANGLES  299 

PROPOSITION  XXXVI.     THEOREM 

532.   Through  a  given  line  not  perpendicular  to  a  plane,  one 
plane  can  be  passed  perpendicular  to  that  plane,  and  only  one. 


Given :  AB  not  J_  to  plane  MN. 

To  Prove :  Through  AB  one  plane  can  be  passed  J_  to  MN, 
and  only  one. 

Proof :  I.     From  P,  any  point  in  AB,  draw  PX  _L  to  MN. 

Through  AB  and  PX  pass  plane  AC. 

Plane  AC  is  J_  to  plane  MN  (525). 

II.  Suppose  another  plane  containing  AB  is  -L  to  plane 
MN. 

Then  the  intersection  AB,  of  these  two  planes,  which  are 
J_  to  plane  MN,  will  be  -L  to  plane  MN  (530). 

But  AB  is  not  -L  to  plane  MN  (Hyp.). 

.*.  there  is  only  one  plane  containing  AB  that  is  _L  to 
plane  MN.  Q.E.D. 

533.  COROLLARY.     The  plane  containing  a  straight  line  and 
its  projection  upon  a  plane  is  perpendicular  to  the  given  plane. 

534.  COROLLARY.     If  a  line  meets  its  projection  on  a  plane, 
any  line  of  the  plane  perpendicular  to  one  of  these  lines  at  their 
intersection  is  perpendicular  to  the  other  also. 

Proof:  Use  fig.  of  505. 

Plane  ABP  is  J_  to  plane  MN  (533). 

A  line  -L  to  plane  ABP  at  B  will  lie  in  plane  MN  (528). 

.-.  this  line  is  J.  to  both  AB  and  PB  (473). 


300 


BOOK   VI.     SOLID   GEOMETRY 


PROPOSITION  XXXVII.     THEOREM 

535.  Between  any  two  straight  lines  not  in  the  same  plane, 
one  and  only  one  common  perpendicular  can  be  drawn,  and 
this  common  perpendicular  is  the  shortest  line  that  can  be 
drawn  between  the  two  lines. 


B 


Given :  Lines  AB  and  CD  not  in  the  same  plane. 

To  Prove :  I.     One  line  can  be  drawn  -L  to  AB  and  CD. 

I 1 .  Only  one  -L  can  be  drawn. 

III.  This  -L  is  the  shortest  line  that  can  be  drawn  be- 
tween AB  and  CD. 

Proof :  I.     At  P,  any  point  in  CD,  draw  EF II  to  AB. 
Pass  plane  JfJV,  containing  CD  and  EF. 
Pass  plane  AH  through  AB  and  _L  to  plane  MN,  intersect- 
ing plane  MN  in  GH,  and  CD  at  L. 
In  plane  AH  draw  EL  _L  to  GH. 

Plane  MN  is  li  to  AB  (482). 

GH  is  II  to  AB  (483). 

EL  is  -L  to  GH  (Const.). 

.-.  EL  is  -L  to  plane  MN  (527). 

.-.  EL  is  _L  to  CD  (473). 

EL  is  J-  to  AB  (64). 


Also 
That  is, 


EL  is  JL  to  both  the  given  lines.    Q.E.D. 


DIHEDRAL   ANGLES  301 

II.  If  another  line  can  be  drawn  J.  to  AB  and  CD,  sup- 
pose SP  is  this  J_.     In  plane  AH  draw  SX  _L  to  GU. 

Then  SX  is  J.  to  plane  MN  (527). 

But  if          SP  is  -L  to  AB,  it  is  -L  to  EF  (64). 

.-.  SP  is  -L  to  plane  MN  (485). 

Thus  there  are  two  J§  from  s  to  plane  M N  (SX  and  SP). 
But  this  is  impossible  (491). 

.-.  there  can  be  no  second  JL  to  these  two  given  lines. 

Q.E.D. 

III.  Suppose  SP  is  any  other  line  between  AB  and  CD. 
Now  EL  is  II  to  sx  (62). 

.-.  EX  is  a  O  (120). 

.-.  RL  =  SX  (124). 

But  sx  <  SP  (504,  I). 

.-.  B£  <  SP  (Ax.  6). 

That  is,  EL  is  shorter  than  any  other  line  between  AB  and 

CD.  Q.E.D. 

Ex.  1.  In  the  figure  of  535  prove  that  a  plane  perpendicular  to  RL 
at  its  midpoint  will  be  parallel  to  AB  and  CD. 

Ex.  2.     Prove,  also,  that  this  plane  will  bisect  SP. 

Ex.  3.  Prove  that  if  CD  is  not  perpendicular  to  EF,  no  plane  can  be 
passed  through  AB,  perpendicular  to  CD. 

Ex.  4.     Tell  how  we  can  construct  one  plane  perpendicular  to  another. 

Ex.  6.  Tell  how  we  can  construct  one  plane  through  a  given  point, 
perpendicular  to  any  two  given  planes. 

Ex.  6.  Tell  how  we  can  construct  a  plane  containing  a  given  line 
and  perpendicular  to  a  given  plane. 

Ex.  7.  Tell,  so  that  a  blind  boy  could  understand,  how  to  draw  a  line 
perpendicular  to  any  two  lines  in  space  (not  in  the  same  plane)., 

Ex.  8.  If  two  planes  are  parallel,  what  is  the  form  of  the  projection 
on  one  plane,  of  a  circle  in  the  other  ? 

Ex.  9.  If  two  planes  are  perpendicular,  what  is  the  form  of  the  pro- 
jection on  one  plane,  of  a  circle  in  the  other? 


302 


BOOK  VI.     SOLID   GEOMETRY 


PROPOSITION   XXXVIII.     THEOREM 

536.   Every  point  in  a  plane  bisecting  a  dihedral  angle  is 
equally  distant  from  the  faces  of  the  angle. 


Given:  Plane  AB,  bisecting  the  dih.  Z  C-BD-E-,   any  point 
P  in  plane  AB  ;  PF  _L  to  face  CB  ;   PH  _L  to  face  DE. 
To  Prove:  PF  =  PH. 

Proof  :   Pass  plane  MN,  containing  PF  and  Pfl,  intersecting 
CB  in  FG,  AB  in  PG,  DE  ill  HG,  BD  at  G. 

Now  plane  MN  is  _L  to  planes  CB  and  DE  (525). 

.  •.  plane  MN  is  _L  to  BD  (531). 

.-.  BG  is  -L  to  PG,  PG,  and  HG  (473). 

Hence  Z  PGF  is  the  plane  Z  of  dih.  Z  A-BD-C  and  Z  PGH 

is  the  plane  Z  of  dih.  Z  A-BD-E  (515). 

These  dih.  z§  are  =  (Hyp.). 

.'.  Z  PGF=  Z  PGff  (521). 

^i  PM?  and  PHG  are  rt.  ^  (473). 

In  the  right  A  PFG  and  PGfl,  P6?  =  PG  (?). 


.'.  PF=PH  (?).     Q.E.D. 

537.  COROLLARY.     Any  point  in  a  dihedral  angle  and  equally 
distant  from  its  faces  is  in  the  plane  bisecting  the  angle. 

To  Prove  :  The  plane  AB,  determined  by  the  point  P  and 
the  edge  BD,  bisects  the  dih.  Z  C-BD-E. 

538.  COROLLARY.     The  locus  of  points  within  a  dihedral  angle 
and  equally  distant  from  its  faces  is  the  plane  bisecting  that 
angle.  (Proof  :  536,  537.) 


ORIGINAL  EXERCISES  303 

ORIGINAL   EXERCISES 

1.  Are    two    planes  perpendicular   to   the   same  plane   necessarily 
parallel? 

2.  A  straight  line  and  a  plane  perpendicular  to  the  same  plane  are 
parallel. 

3.  A  plane  perpendicular  to  a  line  in  another  plane  is  perpendicular 
to  that  plane. 

4.  If  three  planes,  all  perpendicular  to  a  fourth,  intersect  in  three 
lines,  these  lines  are  parallel,  in  pairs. 

6.   If  the  projection  of  any  line  (straight  or  curved)  upon  a  plane  is 
a  straight  line,  the  line  is  entirely  in  one  plane. 

6.  The  angle  between  the  normals  drawn  to  the  faces  of  a  dihedral 
angle  from  a  point  within  the  angle  is  the  supplement  of  the  plane  angle 
of  the  dihedral  angle. 

7.  If  a  line  is  parallel  to  a  plane,  any  plane  perpendicular  to  the  line 
is  perpendicular  also  to  the  plane. 

[Construct  the  projection  of  the  given  line  upon  the  given  plane.] 

8.  What  is  the  locus  of  points  in  space  equally  distant  from  two 
intersecting  planes? 

9.  If  from  any  point  in  a  face  of  a  dihedral  angle, 
a  normal  is  drawn  to  each  face,  the  plane  of  these 
normals  is  perpendicular  to  the  edge  of  the  dihedral 
angle. 

10.  If  from  any  point  in   a  face  of  a  dihedral 
angle,  a  normal  is  drawn  to  each  face,  the  angle  they 
form  is  equal  to  the  plane  angle  of  the  dihedral  angle. 

11.  If  a  line  is  perpendicular  to  a  plane,  any  plane  parallel  to  the 
line  is  also  perpendicular  to  the  plane. 

12.  If  PA    is  a  normal    to    plane   MN,  PB  a  M 
normal  to  plane  ST,  and  BC  a  normal  to  plane  MN,  I 
AC    is    perpendicular   to  RS,   the   intersection   of  /_ 
planes  MAT  and  ST. 

13.  The  plane  perpendicular  to  the  line  that  is  perpendicular  to  two 
lines  in  space,  at  its  middle  point,  bisects  every  straight  line  having  its 
extremities  in  these  lines. 


304 


BOOK  VI.     SOLID   GEOMETRY 


14.  The  plane  perpendicular  to  the  plane  of  an 
angle  and  containing  the  bisector  of  the  angle  is 
the  locus  of  points  equally  distant  from  the  sides  of 
the  angle. 

Proof:  The  J§  from  any  point  in  plane  NR  to 
AB  and  EC  will  have  equal  projections.  (Explain  by  use  of  94.) 

.•.  these  perpendiculars  are  equal  (?). 

16.   What  is  the  locus  of  points  in  space  equally  distant  from  two 
intersecting  lines? 

16.  If  A  P  is  a  normal  to  plane  MN  and  if  angle 
PBC,  in  plane  MN,  is  a  right  angle,  angle  ABC  also 
is  a  right  angle. 

[Prove  EC  is  JL  to  plane  APB.] 

17.  If  AP  is  a  normal  to  plane  MN,  and  Z.PBD, 
in  plane  MN,  is  obtuse,  Z.  ABD  also  is  obtuse. 

Proof :  Take  EC  =  BD ;  prove  PD>PC.    Then  prove  AD  >  A  C,  etc. 

18.  PA   is  perpendicular  to  plane   RS;    AB   and 
PC  are  perpendicular  to  plane  MR.     Prove  EC  per- 
pendicular to  RT. 

19.  If  two  parallel  planes  are  cut  by  a  third  plane, 
the  alternate-interior  dihedral   angles  are  equal;    the 
corresponding  dihedral  angles  are  equal ;  the  alternate- 
exterior  dihedral  angles  are  equal;    the  adjoining  in- 
terior dihedral  angles  are  supplementary. 

20.  State  and  prove  the  converse  theorems  of  those  in  No.  19. 


N 


21.  Construct  a  plane  perpendicular  to  a  given  plane  and  containing 
a  given  line  in  that  plane. 

22.  Construct  a  plane  perpendicular  to  a  given  plane  and  containing 
a  given  line  without  that  plane. 

23.  Construct  through  a  given  point  a  line  which  will  intersect  any 
two  given  lines  in  space. 

Construction :  Pass  a  plane  through  the  point  and  one  of  the  lines. 
This  plane  intersects  the  other  line  at  a  point,  etc. 

24.  To  bisect  a  given  dihedral  angle. 

Construction :  Pass  a  plane  _L  to  the  edge.     Bisect  the  plane  ^  of  the 
given  dihedral,  etc. 


POLYHEDRAL   ANGLES  305 

25.  Construct  a  line  each  of  whose  points  shall  be  equally  distant  from 
the  ends  of  a  given  line  and  also  equally  distant  from  the  faces  of  a  dihe- 
dral angle. 

26.  Find  the  locus  of  points  equally  distant  from  two  points  and 
equally  distant  from  two  intersecting  planes.     Discuss. 

27.  Find  a  point  equally  distant  from  three  given  points  and  equally 
distant  from  two  intersecting  planes.     Is  this  problem  ever  impossible? 
Will  there  ever  be  two  points?     When  will  there  be  only  one  point? 

28.  Find  a  point  equally  distant  from  three  given  points  and  equally 
distant  from  two  intersecting  lines.     Discuss  fully. 

POLYHEDRAL    ANGLES 

539.  If  three  or  more  planes  meet  at  a  point,  they  form  a 
polyhedral  angle.  The  opening  partially  surrounded  by  the 
planes  is  the  polyhedral  angle. 

The  point  common  to  all  the  planes  is  the  vertex. 

The  planes  are  the  faces. 

The  intersections  of  adjacent  faces  are  the  edges. 

The  angles  formed  at  the  vertex,  by  adjacent  edges,  are 
the  face  angles. 

Thus,  V  -  ABODE  is  a  polyhedral  angle;  V  is  the  vertex;  A  V,  BV, 
etc.,  are  edges ;  planes  A  VB,  BVC,  etc.,  are  faces  ;  A  A  VB,  B  VC,  etc., 
are  face  angles. 

Vy 


540.    A  plane  section  of  a  polyhedral  angle  is  the  plane 

figure  bounded  by  the  intersections  of  all  the  faces  by  a  plane. 

Polygon  LMNOP  is  a  plane  section  of  polyhedral  angle  V— ABODE. 

A  convex  polyhedral  angle  is  one  whose  plane  sections  are 
all  convex. 


306 


BOOK   VI.     SOLID   GEOMETRY 


B 


EQUAL  POLYHEDRAL 
ANGLES 


VERTICAL 

POLYHEDRAL 

ANGLES 


VERTICAL 

DIHEDRAL 

ANGLES 


SYMMETRICAL 

POLYHEDRAL 

ANGLES 


641.  Two  polyhedral  angles  are  equal  if  they  can  be  made 
to  coincide  in  all  particulars.  That  is,  if  two  polyhe- 
dral angles  are  equal,  their  homologous  dihedral  angles  are 
equal;  their  homologous  face  angles  are  equal,  and  they 
are  arranged  in  the  same  order.  The  length  of  the  edges 
or  the  extent  of  the  faces  does  not  affect  the  size  of  the 
angle. 

Two  polyhedral  angles  are  vertical  if  the  edges  of  one  are 
the  prolongations  of  the  edges  of  the  other. 

Two  polyhedral  angles  are  symmetrical  if  all  the  parts  of 
one  are  equal  to  the  corresponding  parts  of  the  other,  but 
arranged  in  opposite  order. 

NOTE.  It  is  apparent  from  the  definitions  that  equal  polyhedral 
angles  are  mutually  equiangular  as  to  the  face  angles  and  as  to  the  dihe- 
dral angles. 

Vertical  polyhedral  angles  are  mutually  equiangular  as  to  their  face 
angles  and  as  to  their  dihedral  angles,  but  the  order  is  reversed. 

Symmetrical  polyhedral  angles  are  also  mutually  equiangular  as  to 
their  face  angles  and  as  to  their  dihedral  angles,  but  the  order  is  reversed. 
Thus,  if  one  follows  around  the  polygon  A'D'  in  alphabetical  order, 
he  is  moving  as  the  hands  of  a  clock,  if  the  eye  is  at  the  vertex  0'  ; 
but  if  he  follows  around  AD  alphabetically,  he  is  moving  in  a  direction 
opposite  to  the  motion  of  the  hands  of  a  clock,  if  the  eye  is  at  the 
vertex  O.  Hence,  it  is  apparent  that,  in  general,  symmetrical  polyhedral 
angles  cannot  be  made  to  coincide. 


POLYHEDRAL    ANGLES  307 

642.  A  trihedral  angle  is  a  polyhedral  angle  having  three 
and  only  three  faces. 

A  trihedral  angle  is  rectangular  if  it  contains  a  right  di- 
hedral angle  ;  birectangular  if  it  contains  two  right  dihedral 
angles;  trirectangular  if  it  contains  three  right  dihedral  angles. 

A  trihedral  angle  is  isosceles  if  two  of  its  face  angles  are 
equal. 

PRELIMINARY  THEOREMS 
543.   Two  vertical  polyhedral  angles  are  symmetrical. 

Proof  :  Their  homologous  face  angles  are  equal  and  ar- 
ranged in  reverse  order,  and  their  homologous  dihedral  angles 
are  equal  and  arranged  in  reverse  order. 

.-.  they  are  symmetrical  (Def.  541). 

644.  If  two  polyhedral  angles  are  symmetrical,  the  vertical 
polyhedral  angle  of  the  one  is  equal  to  the  other. 

Because  the  corresponding  parts  are  equal  and  they  are 
arranged  in  the  same  order. 

645.  Provided  two  trihedral  angles  have  their  parts  arranged 
in  the  same  order,  they  are  equal  :  * 

I.  If  two  face  angles  and  the  included  dihedral  angle  of  one 
are  equal  respectively  to   two  face  angles  and  the  included 
dihedral  angle  of  the  other. 

II.  If  a  face  angle  and  the  two  dihedral  angles  adjoining  it 
of  the  one  are  equal  respectively  to  a  face  angle  and  the  two 
dihedral  angles  adjoining  it,  of  the  other. 

Proof :  By  method  of  superposition,  as  in  plane  A. 

Ex.  Illustrate  a  birectanmilar  trihedral  angle  by  means  of  an  open 
book  standing  on  a  table.  Similarly  illustrate  a  trirectangular  trihedral 
angle.  Similarly  use  two  closed  books  to  illustrate  a  rectangular  trihe- 
dral angle. 

ROBBINS'8    NEW   SOLID   GEOM. — 6 


308 


BOOK   VI.     SOLID   GEOMETRY 


PROPOSITION   XXXIX.     THEOREM 

546.  Provided  two  trihedral  angles  have  their  parts  arranged 
in  the  same  order,  they  are  equal,  if  the  three  face  angles  of 
one  are  equal  respectively  to  the  three  face  angles  of  the  other. 


Given :  Trih.  A  o  and  o' ; 


Z  BOG  = 

Z  COA  =  Z  C'0'^l'. 

To  Prove:  Trih.  Z  o  =  trih.  Z  o',  that  is,  dih.  Z  o^i  =  dih. 
Z  C/.4',  etc. 

Proof:  Take  OA  =  O7?  =  OC  =  o'^i'  =  O'B'  =  o'c'. 

Draw  4£,  BC,  AC,  A'B',  B'C',  A'C'. 

Take,  on  edges  AO  and  A'o',  AP  =  A'P'  and  in  face  A  OB 
draw  PD  J_  to  .40. 

Z  O.4.B  is  acute  (A  AOB  is  isosceles).     .'.  PD  will  meet  AB. 

In  face  AOC  draw  PEA.  to  .40,  meeting  .4(7  at  E. 

Draw  DJ£.     Similarly  draw  P'D',  P'E',  D'E'. 

Now  AEPD  and  E1  P'D1  are  the  plane  z§  of  the  dihedral  A 
AO  and  .4'o'  (515). 

To  prove  that  these  A  are  equal  requires  the  proof  that 
eight  pairs  of  A  are  equal. 

.-.AB  =  A'B' 

(1)  AO^B  =  A  O'A'B'     (Explain). 

(2)  A  OBC  =  A  O'B'C'     (Explain). 


(3)  AOAC  =  A  o'A'cf     (Explain). 


BC  =  B'C' 
AC  =  A'C' 


Z  OAB  =  z  O'A'B', 


POLYHEDRAL   ANGLES  309 


(4)  AAPD  =  AA'PfD'    (Explain). 

ST\         i     •      N  A Ei  —  A  Ei 

(Explain).  pj^p/^etc. 

(6)  AABC  =  AA'B'C'     (Explain).   .-.  z^CAB  =  Z  CfAr  Br    (?). 

(7)  AAED  =  AA'E'D'    (Explain).       .'.ED=E'D'         (?). 

(8)  A PED  =  A  P'E'D'     (Explain).  .-.  Z.EPD  =  ^EfPrDr     (?). 
Hence  dih.  Z^O  =  dih.  ZA'O'  (520). 

Similarly,  one  may  prove  the  other  pairs  of  homologous 
dihedral  angles  equal. 

.-.  trihedral  Z  o  =  trihedral  Z  o'  (541). 

Q.E.D. 

PROPOSITION  XL.     THEOREM 

547.  Provided  two  trihedral  angles  have  then*  parts  arranged 
hi  reverse  order,  they  are  symmetrical : 

I.  If  two  face  angles  and  the  included  dihedral  angle  of  one 
are  equal  respectively  to  two  face   angles  and  the  included 
dihedral  angle  of  the  other. 

II.  If  a  face  angle  and  the  two  dihedral  angles  adjoining  it 
of  the  one  are  equal  respectively  to  a  face  angle  and  the  two 
dihedral  angles  adjoining  it  of  the  other. 

III.  If  the  three  face  angles  of  one  are  equal  respectively  to 
the  three  face  angles  of  the  other. 

Proof :  In  each  case  construct  a  third  trihedral  Z  symmet- 
rical to  the  first.  This  third  figure  will  have  its  parts  =  to  the 
parts  of  the  second,  and  arranged  in  the  same  order  (Def .  541). 

.-.  the  third  =  the  second  (545  and  546). 

.*.  the  first  is  symmetrical  to  the  second.      (Ax.  6.) 

Q.E.D. 

Ex.  The  three  planes  bisecting  the  three  dihedral  angles  of  a  tri- 
hedral angle  intersect  in  a  straight  line. 


310  BOOK   VI.     SOLID   GEOMETRY 

PROPOSITION  XLI.     THEOREM 

548.   The  sum  of  any  two  face  angles  of  a  trihedral  angle  is 
greater  than  the  third  face  angle. 


Given:  Trih.  Z.O-RST  .in   which   face   angle   EOT  is  the 
greatest. 

To  Prove :  Z  EOS  +  Z  SOT  >  Z  ROT. 

Proof:  Construct,  in  face  EOT,  Z ROD  =  Z.EOS. 
Take  OD—  OB;  draw  ADC,  meeting  OT  at  C.     Draw  AB  and 
BC. 

A  AOD  ^  A  AOB  (Explain). 

.'.AB  =  AD  (?). 

Now  AB  +  BC>  AD  +  DC  (Ax.  12). 

But  AB  =AD  (?). 

Subtracting,  BO  DC  (Ax.  7). 

Now  OB  =  OD  (?),  OC  —  OC  (?)  and  BC  >  DC  (Just  proved). 
. •.  Z  BOC  >  Z DOC  (92). 

But  /.AOB =  Z^OD  (?). 

Adding,  Z  ^iOB  +  Z  BOC  >  Z  ^loc  (Ax.  7). 

That  is,     Z  EOS  +  Z  sor  >  Z  HOT  (Ax.  6). 


Ex.  1.  Prove  theorem  of  548  for  the  case  of  an  isosceles  trihedral 
angle. 

Ex.  2.  The  three  planes  containing  the  three  bisectors  of  the  three 
face  angles  of  a  trihedral  angle  and  perpendicular  to  those  faces  intersect 
in  a  straight  line. 


POLYHEDRAL   ANGLES  311 

PROPOSITION  XLII.     THEOREM 

549.  The  sum  of  the  face  angles  of  any  polyhedral  angle 
is  less  than  four  right  angles  or  360°. 

O. 


Given :  Polyhedral  Z  O,  having  n  faces. 

To  Prove :  The  sum  of  the  face  A  at  O  <  4  rt.  A  or  360°. 

Proof:  Pass  a  plane  AD,  intersecting  all  the  faces,  and  the 
edges  at  A,  B,  C,  etc. 

In  this  section  take  any  point  X  and  join  X  to  all  the 
vertices  of  the  polygon. 

(1)  There  are  n  face  A  having  their  vertices  at  O  (Hyp.). 

(2)  There  are  n  base  A  having  their  vertices  at  X 

(Const.). 

(3)  The  sum  of  the  A  of  the  face  A  =  2n  rt.  A        (104). 

(4)  The  sum  of  the  A  of  the  base  A  =  2w  rt.  A       (104). 

(5)  .*.  the  sum  of  the  A  of  the  face  A  =  the  sum  of  the 
A  of  the  base  A  (Ax.  1). 

Now  Z  OAE  +  Z  OAB  >  Z  EAB  (548). 

And  Z  OB  A  -f  Z  OBC  >  Z  ABC        (?),  etc.,  etc. 

Adding,  the  sum  of  the  base  A  of  the  face  A  >  the  sum  of 
the  base  A  of  the  base  A  (Ax.  8). 

Subtracting  this  inequality  from  equation  (5)  above,  the 
sum  of  the  face  A  at  O  <  the  sum  of  the  A  at  X  (Ax.  9). 

But  the  suni  of  all  the  A  at  X  =  4  rt.  A  (47). 

.-.  the  sum  of  the  face  A  at  O  <  4  rt.  A,  or  360°      (Ax.  6). 

Q.E.D. 


312  BOOK  VI.     SOLID   GEOMETRY 

ORIGINAL  EXERCISES 

1.  In  the  figure  of  549,  as  the  vertex  O  approaches  the  base,  does 
the  sum  of  the  face  angles  at  O  increase  or  decrease  ?     What  limit  does 
this  sum  approach  ?     Does  the  sum  ever  become  equal  to  this  limit? 

2.  Can   a  polyhedral   angle   have    for   its   faces   three   equilateral 
triangles?  four?  five?  six?  seven? 

3.  Can  a  polyhedral  angle  have  for  its  faces  four  squares?   five? 
three? 

4.  What  other  regular  polygons  can  be  used  for  the  faces  of   a 
polyhedral  angle  ? 

6.  If  two  face  angles  of  a  trihedral  angle  are 
equal,  the  dihedral  angles  opposite  them  are  equal. 

Given:  /.RVS  =  £SVT. 

To  Prove :  Dih.  Z  R  V  =  dih.  Z  TV. 

Proof:  Pass  plane  S  VX  bisecting  dih.  ZSV.  Prove 
trih.  A  V-RSX  and  V-TSX  are  sym.  by  (547,  1). 

6.  An  isosceles  trihedral  angle  and  its  symmetri- 
cal trihedral  angle  are  equal. 

7.  Find  the  locus  of  points  equally  distant  from  the  three  faces  of  a 
trihedral  angle. 

8.  Find  the  locus  of  points  equally  distant  from  the  three  edges  of  a 
trihedral  angle. 

9.  If  the  three  face  angles  of  a  trihedral  angle  are  equal,  the  three 
dihedral  angles  also  are  equal. 

10.  If  the  three  face  angles  of  a  trihedral  angle  are  right  angles,  the 
three  dihedral  angles  also  are  right  angles. 

11.  In  any  trihedral  angle  the  greatest  dihedral  angle  has  the  greatest 
face  opposite  it. 

12.  If  the  edges  of  one  trihedral  angle  are  perpendicular  to  the  faces 
of  a  second  trihedral  angle,  then  the  edges  of  the  second  are  perpendicu- 
lar to  the  faces  of  the  first. 

13.  Construct,  through  a  given  point,  a  plane  which  shall  make,  with 
the  faces  of  a  polyhedral  angle  having  four  faces,  a  section  that  is  a 
parallelogram. 

Construction:  Extend  one  pair  of  opp.  faces  to  obtain  their  line  of 
intersection.  Similarly  extend  the  other  pair.  Any  plane  section  II  to 
these  lines  will  be  a  O.  [Explain.] 


BOOK  VII 


POLYHEDRONS 

650.    A  polyhedron  is  a  solid  bounded  by  planes. 

The  edges  of  a  polyhedron  are  the  intersections  of  the 
bounding  planes. 

The  faces  are  the  portions  of  the  bounding  planes  included 
by  the  edges. 

The  vertices  are  the  intersections  of  the  edges. 

The  diagonal  of  a  polyhedron  is  a  straight  line  joining  two 
vertices  not  in  the  same  face. 


POLY-    TETRA-   HEXAHEDRON   OCTA-     DODECA-    ICOSA- 
HEDRON    HEDRON     CUBE     HEDRON    HEDRON   HEDRON 

651.    A  tetrahedron  is  a  polyhedron  having  four  faces. 
A  hexahedron  is  a  polyhedron  having  six  faces. 
An  octahedron  is  a  polyhedron  having  eight  faces. 
A  dodecahedron  is  a  polyhedron  having  twelve  faces. 
An  icosahedron  is  a  polyhedron  having  twenty  faces. 

552.    A  polyhedron  is  convex  if  the  section  made  by  every 
plane  is  a  convex  polygon. 

Only  convex  polyhedrons  are  considered  in  this  book. 

313 


314 


BOOK   VII.     SOLID   GEOMETRY 


PRISMS 


553.  A  prism  is  a  polyhedron  two  of  whose  opposite  faces 
are  congruent  polygons  in  parallel  planes,  and  whose  other 
faces  are  all  parallelograms. 

The  bases  of  a  prism  are  the  congruent  parallel  polygons. 

The  lateral  faces  of  a  prism  are  the  parallelograms. 

The  lateral  edges  of  a  prism  are  the  intersections  of  the 
lateral  faces. 

The  lateral  area  of  a  prism  is  the  sum  of  the  areas  of  the 
lateral  faces. 

The  total  area  of  a  prism  is  the  sum  of  the  lateral  area  and 
the  areas  of  the  bases. 

The  altitude  of  a  prism  is  the  perpendicular  distance  be- 
tween the  planes  of  the  bases. 

A  triangular  prism  is  a  prism  whose  bases  are  triangles. 


PRISM    TRIANGULAR    REGULAR      OBLIQUE  PRISMS         RIGHT  SECTION 
PRISM  PRISM  TRUNCATED  PRISMS 

554.  A  right  prism  is  a  prism  whose  lateral  edges  are  per- 
pendicular to  the  planes  of  the  bases. 

A  regular  prism  is  a  right  prism  whose  bases  are  regular 
polygons. 

An  oblique  prism  is  a  prism  whose  lateral  edges  are  not  per- 
pendicular to  the  planes  of  the  bases. 

A  truncated  prism  is  the  portion  of  a  prism  included 
between  the  base  and  a  plane  not  parallel  to  the  base. 


PRISMS  315 

A  right  section  of  a  prism  is  the  section  made  by  a  plane 
perpendicular  to  the  lateral  edges  of  the  prism. 


PARALLELEPIPED  RIGHT  CUBE  RECTANGULAR 

PARALLELEPIPED  PARALLELEPIPED 

555.  A  parallelepiped  is  a  prism  whose  bases  are  parallelo- 
grams. 

A  right   parallelepiped   is   a   parallelepiped  whose   lateral 
edges  are  perpendicular  to  the  planes  of  the  bases. 

A  rectangular  parallelepiped  is  a  right  parallelepiped  whose 
bases  are  rectangles. 

An  oblique  parallelepiped  is  a  parallelepiped  whose  lateral 
edges  are  not  perpendicular  to  the  planes  of  the  bases. 

A  cube  is  a  rectangular  parallelepiped  whose  six  faces  are 
squares. 

556.  The  unit  of  volume  is  a  cube  whose  edges  are  each  a 
unit  of  length.     The  volume  of  a  solid  is  the  number  of  units 
of  volume  it  contains.     The  volume  of  a  solid  is  the  ratio  of 
that  solid  to  the  unit 'of  volume. 

The  three  edges  of  a  rectangular  parallelepiped  meeting  at 
any  vertex  are  the  dimensions  of  the  parallelepiped. 

Equal  solids  are  solids  that  have  equal  volumes. 
Congruent  solids  are  solids  that  can  be  made  to  coincide. 


316  BOOK   VII.     SOLID   GEOMETRY 

Ex.  What  is  the  base  of  a  rectangular  parallelepiped  V  of  a  right 
parallelepiped  ?  of  an  oblique  parallelepiped  ?  Illustrate,  by  removing  the 
cover  and  bottom  of  an  ordinary  cardboard  box  and  distorting  the  shape 
of  the  frame  that  remains,  the  three  kinds  of  parallelepipeds. 

NOTE.  The  space  that  is  bounded  by  the  surfaces  of  a  solid,  independ- 
ent of  the  solid,  is  called  a  geometrical  solid. 

That  is,  if  a  material  or  physical  body  should  occupy  a  certain  position 
and  then  be  removed  elsewhere,  there  is  a  definite  portion  of  space  that 
is  the  same  shape  and  size  as  the  solid,  and  can  be  conceived  as  bounded 
by  exactly  the  same  surfaces  as  bounded  the  solid  when  in  that  original 
position.  In  order  that  we  may  pass  planes  and  draw  lines  through 
solids,  and  superpose  one  solid  upon  another,  it  is  convenient  in  studying 
the  properties  of  solids  to  consider  them  usually  as  geometric  solids, — 
the  material  body  being  removed  for  the  time. 

PRELIMINARY  THEOREMS 

657.  THEOREM.     The  lateral  edges  of  a  prism  are  equal.  (?.) 

658.  THEOREM.     Any  two    lateral    edges   of    a  prism  are 
parallel.  (495.) 

559.  THEOREM.    Any  lateral  edge  of  a  right  prism  equals  the 
altitude.  (509.) 

560.  THEOREM.     The  lateral  faces  of  a  right  prism  are  per- 
pendicular to  the  bases.  (525.) 

561.  THEOREM.     The    lateral  faces  of   a  right  prism  are 
rectangles.  (Def.  554.) 

562.  THEOREM.     The  faces  and  bases  of    a   rectangular 
parallelepiped  are  rectangles.  (Def.  555.) 

663.  THEOREM.  All  the  faces  of  any  parallelepiped  are 
parallelograms.  (? ) 

564.  AXIOM.     A  polyhedron  cannot  have  fewer  than  four 
faces. 

565.  AXIOM.     A  polyhedron  cannot  have  fewer  than  three 
faces  at  each  vertex. 


PRISMS 


317 


THEOREMS   AND    DEMONSTRATIONS 

PROPOSITION  I.     THEOREM 

566.   The  sections  of  a  prism  made  by  parallel  planes  cutting 
all  the  lateral  edges  are  congruent  polygons. 


Given:     Prism  AB\   II  sections  CF and  C'F1. 
To  Prove:     Polygon  CF^ polygon  C'F'. 

Proof:     CD  is  II  to  C'D',  DE  is  II  to  D'E',  etc. 

.'.  CD',  DE',  EF',  etc.  are  £17 
.*.  CD  =  C'D',  DE=  D'E',  EF  =  E'F',  etc. 
Z  GCD  =  z.  G'C'D',  Z  CDE  =  Z.C'D'E',  etc. 
.*.  Polygon  CF^  polygon 


(484). 

(?)• 

(124). 

(499). 
(150). 

Q.E.D. 


Ex.  1.     All  right  sections  of  a  prism  are  equal. 

Ex.  2.     Any  section  of   a  parallelepiped  made   by  a  plane  cutting 
two  pairs  of  opposite  faces  is  a  parallelogram. 

Ex.  3.     How  many  edges  has  a  cube?  how  many  vertices?  how  many 
dihedral  angles?  how  many  trihedral  angles? 

Ex.  4.     In  the  figure  of  566,  if  DE  is  the  altitude  of  one  of  the  lateral 
faces,  how  could  one  express  the  area  of  that  face  ? 

Ex.6.     In  566,  is  CF'  a  prism  ?     Why?    Is  CB  a  prism?    Why?     Is 
A F  a  prism  ?    Why  ?    What  name  is  given  CB  ? 


818 


BOOK  VII.     SOLID   GEOMETRY 


PROPOSITION  II.     THEOREM 

667.   The  opposite  faces  of   a  parallelepiped  are  congruent 

and  parallel. 

H G 


Given:     (?). 

To  Prove:     Face  ^.F^and  II  to  face  DG. 

Proof  :  Faces  AF  and  DG  are  ZI7 

AB  =  DC,  AE  =  DH 
AB  is  II  to  DC  and  AE  is  II  to  DH 


.'.  face  AF^  face  DG 
Also  face  AF  is  II  to  face  DG 


PROPOSITION  III.     THEOREM 


(124). 

(?)• 

(499). 

(133). 
(499). 

Q.E.D. 


568.   The  lateral  area  of  a  prism  is  equal  to  the  product  of  a 
lateral  edge  by  the  perimeter  of  a  right  section. 


Given  :  Prism  RU'  ;  edge  =  E  ;  right  section  AD. 


PRISMS  319 

To  Prove  :   Lateral  area  of  EU1  =  E  x  perimeter  of  AD. 

Proof  :  AB  is  J_  to  RRf,  EC  is  _L  to  S81,  etc.  (473). 

Area  O  #,s'  =  ^  .  AB  (359). 

Area  O  sr'  =  E  •  EC  (?). 

Area  O  ru7  =  E  •  CD  (?). 

etc.         etc.         Adding, 

The  lateral  area  =  E  •  (AE  +  EC  +  CD+  etc.)  (Ax.  2). 

=  E  •  perimeter  of  rt.  sect.  (Ax.  6). 

Q.E.D. 

569.  COROLLARY.    The  lateral  area  of  a  right  prism  is  equal 
to  the  product  of  its  altitude  and  the  perimeter  of  its  base. 

L  =  H  -  Pr. 

(Where  L  =  lateral  area,  H  =  altitude,  and  Pr  =  perimeter 
of  base,  of  a  right  prism.) 

570.  COROLLARY.  T  =  L  +  2  B. 

(Where  T  =  total  area,  and  E  =  area  of  base.) 


Ex.  1.  Any  section  of  a  parallelepiped  made  by  a  plane  parallel  to  any 
edge  is  a  parallelogram. 

Ex.  2.  The  sum  of  the  face  angles  at  all  the  vertices  of  any  parallele- 
piped is  equal  to  24  right  angles. 

Ex.  3.  The  sum  of  the  plane  angles  of  all  the  dihedral  angles  of  any 
parallelepiped  is  equal  to  12  right  angles. 

Proof :  Pass  three  planes  JL  to  three  intersecting  edges.  Prove  these 
sections  £i7  whose  A  are  the  plane  angles  of  the  dihedral  angles,  etc. 

Ex.  4.  Find  the  lateral  area  of  a  right  prism  whose  altitude  is  8  ft. 
and  each  side  of  whase  triangular  base  is  5  ft. 

Ex.  5.  Find  the  total  area  of  a  regular  prism  whose  base  is  a  regular 
hexagon,  10  in.  on  a  side,  if  the  altitude  of  the  prism  is  15  in. 

Ex.  6.  Find  the  lateral  area  of  a  prism  whose  edge  is  12  in.  and  whose 
right  section  is  a  pentagon,  the  sides  of  which  are  3  in.,  5  in.,  6  in.,  9  in., 
and  11  in. 


320  BOOK  VII.     SOLID   GEOMETRY 

PROPOSITION  IV.     THEOREM 

571.  Two  prisms  are  congruent  if  three  faces  including  a 
trihedral  angle  of  one  are  congruent,  respectively,  to  three 
faces  including  a  trihedral  angle  of  the  other,  and  similarly 
placed. 

Given:  Prisms  AO  and 
A'o'-,  face  AM  ^  face  AfMr ; 
face  AP  ^  face  A'P'\  face 
AD  3*  face  A'D'. 

To  Prove : 

Prism  AO  ^  prism  ArOf. 

Proof :     The  three  face  A  at  A  are  respectively  =  to  the 
three  face  A  at  Af  (27). 

.  •.  trih.  Z  A  =  trih.  Z.  Af  (546). 

Superpose  prism  AO  upon  prism  A'o',  making  the  equal 
trihedral  A  A  and  A!  coincide. 

Face  AD  coincides  with  face  AfDf,  face  AM  with  A'M',  face 
AP  with  A'Pf  (They  are  ^ by  hyp.). 

That  is,  point  L  falls  on  L1;  M  on  M1 ';  and  P  on  P'. 
.*.  the  plane  £O  falls  upon  the  plane  L'O'  (477). 

Polygon  LO^  polygon  LrOf  (Ax.  1). 

.-.  these  bases  coincide  (?). 

Similarly  face  BN  coincides  with  BfNf,  CO  with  c'o',  etc. 

.•.  the  prisms  are  ^    (Def.  556).     Q.E.D. 

572.  COROLLARY.    Two  right  prisms  are  congruent  if  they 
have  congruent  bases  and  equal  altitudes.  (Explain.) 

573.  COROLLARY.    Two  truncated  prisms  are  congruent  if 
three  faces  including  a  trihedral  angle  of  one  are  congruent, 
respectively,  to  three  faces  including  a  trihedral  angle  of  the 
other,  and  are  similarly  placed.  (Explain.) 


PRISMS 
PROPOSITION  V.     THEOREM 


321 


574.  An  oblique  prism  is  equal  to  a  right  prism  whose  base 
is  a  right  section  of  the  oblique  prism,  and  whose  altitude  is 
equal  to  the  lateral  edge  of  the  oblique  prism. 


Given:  Oblique  prism  AC1 ;  right  prism  PNr  whose  base  is 
PN,  a  right  section  of  AC1 ',  and  whose  altitude  PP'  =  edge  EEf. 
To  Prove:  Oblique  prism  ACf  =  right  prism  PNr. 
Proof :  Edge  EEf  =  PPr  (Hyp.). 

Subtract  PEr  from  each,  and  EP  =  EfPf  (Ax.  2). 

Likewise  AL  =  ArLr,  BM  =  B'M',  CN  ^  CfNf,  etc. 

(1)  Face  AC^ face  A'c'  (553). 

(2)  In  faces  AP  and  A'Pf,  EP  =  EfPr,  AL=  ArLf    (Ax.  2). 
Also  AE=A'E',  PL  =  P'L'  (124). 
That  is,  face  AP  and  face  A'P'  are  mutually  equilateral. 
Also             Z  EAL  =  Z  E'A'L',  Z  P.^1  =  Z 

z  ^LLP  =  z  A'L'P',  z  ^PL  =  Z 
That  is,  faces  AP  and  ^'p'  are  mutually  equiangular. 

.  •.  face  AP  ^  face  A'P'  (150). 

(3)  Similarly,  face  AM  ^face  A'M'. 

.-.  truncated  prism  AN ^ truncated  prism  A' N?          (573). 

Now,        add,  solid  PC'  =  solid  PC1  (Iden.). 

Oblique  prism  AC1  =  right  prism  PN1   (Ax.  2).  Q.E.D. 

Ex.     Prove,  in  the  figure  of  574,  that  truncated  prism  A  N  is  congruent 
to  truncated  prism  A' N',  by  the  method  of  superposition. 


(67). 


322  BOOK   VII.     SOLID   GEOMETRY 

PROPOSITION  VI.     THEOREM 

675.  The  plane  containing  two  diagonally  opposite  edges  of 
a  parallelepiped  divides  the  parallelepiped  into  two  equal  tri- 
angular prisms. 


Given:  Parallelepiped  BH  and  plane  AG  containing  the 
opposite  edges  AE  and  CG. 

To  Prove :  Prism  ABC-  F=  prism  ADC  -  H. 
Proof:  Pass  a  right  section  RSTV  intersecting  the  given 
plane  in  RT. 

Face  AF  is  II  to  face  DG  (?). 

.-.  RS  is  II  to  VT  (484). 

Also  RV  is  II  to  8T  (?). 

.-.  RSTV  is  a  O  (?). 

.'.  ARST^A  EVT  (126)- 

Prism  ABC-F=&  right  prism  whose  base  is  RST  and  whose 

altitude  =  EA  (574). 

Prism  ADC-H  =  a  right  prism  whose  base  is  RVT  and  whose 

altitude  =  EA  (?). 

But  these  imaginary  right  prisms  are  congruent         (572). 

.-.  prism  ABC-F=  prism  ADC-H  (Ax.  1). 

Q.E.D. 

Ex.  1.  The  section  of  a  parallelepiped  made  by  a  plane  containing  two 
diagonally  opposite  edges,  is  a  parallelogram. 

Ex.  2.  Are  the  two  triangular  prisms  in  the  diagram  of  575  congruent  ? 
Why? 


PARALLELEPIPEDS 


PROPOSITION  VII.     THEOREM 


323 


576.  Two  rectangular  parallelepipeds  having  congruent  bases 
are  to  each  other  as  their  altitudes. 


Given :  Rectangular  parallelepipeds    P  and    Q,    having  ^ 
bases,  and  their  altitudes  AB  and  CD,  respectively. 

To  Prove :  P  :  Q  =  AB  :  CD. 
Proof :  I.    If  the  altitudes  are  commensurable. 
(Consult  524. ) 


II.    If  the  altitudes  are  incommensurable. 
There  does  not  exist  a  common  unit 


(225). 


Suppose  AB  divided  into  equal  parts.     Apply  one  of  these 
as  a  unit  of  measure  to  CD.     There  will  be  a  remainder,  DX 


Pass  a   plane  XT,  through  X  and  II  to  the  base. 
Now  P:CY=AB:CX 

Indefinitely  increase,  etc.  —  as  in  524. 

ROBBINGS   NEW   SOLID  OEOM.  —  6 


(?). 


324 


BOOK    VII.     SOLID   GEOMETRY 


577.  COROLLARY.  Two  rectangular  parallelepipeds  having 
two  dimensions  of  the  one  equal  respectively  to  two  dimen- 
sions of  the  other,  are  to  each  other  as  their  third  dimension. 

The  faces  having  the  sides  of  one  equal  to  the  sides  of  the  other,  re- 
spectively, may  be  considered  the  bases  and  the  third  dimensions  the 
altitudes.  Thus  this  statement  is  the  same  as  576. 


PROPOSITION  VIII.     THEOREM 

678.  Two  rectangular  parallelepipeds  having  equal  altitudes 
are  to  each  other  as  their  bases. 


Hi 


Given:  Rectangular  parallelepipeds  R  and  5,  having  the 
same  altitude  H  ;  and  other  dimensions  a,  5,  and  <?,  c?,  respec- 
tively. 

To  Prove:    *  =  ^. 

S         C  '  d 

Proof  :  Construct  a  third  rectangular  parallelepiped,  having 
altitude  =  to  £T,  another  dimension  =  to  a,  a  third  =  to  d. 


Multiplying, 


R      b       ,  T     a 

—  =  —  and  —  =  — 

T      d          S      c 
E  __a  •  b 

S~  c  -d 


(577). 


(Ax.  3). 

Q.E.D. 

579.  COROLLARY.  Two  rectangular  parallelepipeds  having 
one  dimension  in  common,  are  to  each  other  as  the  products 
of  the  other  dimensions. 


PARALLELEPIPEDS 


325 


PROPOSITION  IX.     THEOREM 

580.    THEOREM.     Any  two   rectangular  parallelepipeds   are 
to  each  other  as  the  products  of  their  three  dimensions. 


N 


a 


Given:  Rectangular   parallelepipeds   L  and  3f,  whose   di- 
mensions are  «,  6,  Zf,  and  a1 ',  6',  H1 ',  respectively. 


To  Prove :     -  = 


Proof:  Construct  JV,  whose  dimensions  are  a,  5,  fl'. 

Then  •-  =  —,  <577)' 

N      H 
N       a  •  b 


And 
Multiplying, 


M     a'  •  bf 
L         a  •  b  •  H 


M      a'  >b'  .H' 


(579). 
(Ax.  3). 

Q.E.D. 


Ex.  1.     In  the  above  diagram,  if  a  =  6  in.,  b  =  6  in.,  and  H  =  7  in., 
find  the  length  of  the  diagonal  of  L. 

Ex.  2.     Find  the  length  of  the  diagonal  of  a  room  a  ft.  long,  b  ft.  wide, 
and  c  ft.  high. 

Ex.  3.     Show  that  the  four  diagonals  of  a  rectangular  parallelepiped 
are  all  equal. 

Ex.  4.     Find  the  diagonal  of  a  cube  whose  edge  is  4  in.     Find  the  edge 
of  a  cube  whose  diagonal  is  10  in. 


326 


BOOK   VII.     SOLID   GEOMETRY 


PROPOSITION  X.     THEOREM 

581.    The  volume  of  a  rectangular  parallelepiped  is  equal  to 
the  product  of  its  three  dimensions. 


Given:  (?).        To  Prove:  (?). 

Proof :  Let  U  be  a  unit  of  vol. 

P  =  a  -  b  -  H 
U~  1-1.1 


=  a  •  b  •  H 


But 


-  =  vol.  of  P 
U 

vol.  of  P  =  a  «  b  -  H 


(580). 
(556). 


(Ax.  1). 

Q.E.D. 

582.     COROLLARY.    The  volume  of  a  rectangular  parallele- 
piped is  equal  to  the  product  of  its  base  by  its  altitude. 


V  =  B  •  H. 


(See  581.) 


583.  COROLLARY.  The  volume  of  a  cube  is  equal  to  the 
cube  of  its  edge. 

Ex.  1.  A  rectangular  tank  is  56  in.  long,  44  in.  wide,  and  60  in.  deep 
(inside).  How  many  gallons  will  it  hold?  [231  cu.  in.  =  1  gal.] 

Ex.  2.  Three  persons  measured  the  above  tank  inaccurately.  A  found 
it  to  be  57  in.  x  44  in.  x  60  in. ;  B,  56  in.  x  45  in.  x  60  in. ;  and  C,  56 
in.  x  44  in.  x  61  in.,  —  each  recording  two  dimensions  correctly,  and  one, 
1  in.  too  great.  Whose  error  was  most  serious,  judging  by  the  capacity 
of  the  tank?  Which  dimensions  should  be  most  carefully  measured? 

Ex.  3.  What  is  the  volume  of  a  cube  6  in.  on  each  edge?  What  is 
the  edge  of  a  cube  having  double  the  volume?  What  happens  to  the 
volume  of  a  cube  if  we  double  each  edge  ?  if  we  halve  each  edge  ? 


PARALLELEPIPEDS  327 

PROPOSITION  XI.     THEOREM 

584.     The  volume  of  any  parallelepiped  is  equal  to  the  prod- 
uct of  its  base  by  its  altitude. 


Given :  Parallelepiped  #,  whose  base  =  B  and  alt.  =  H. 

To  Prove :  Volume  of  E  —  B  -  H. 

Proof :  Prolong  the  edge  AD  and  all  edges  II  to  AD. 

On  the  prolongation  of  AD,  take  EF=  to  AD. 

Through  E  and  F  pass  planes  EG  and  .FT,  _L  to  EF,  form- 
ing the  right  parallelepiped  S. 

Again,  prolong  FJ  and  all  the  edges  II  to  FJ. 

On  the  prolongation  of  FJ,  take  KL  =  to  FJ. 

Through  K  and  L  pass  planes  KM  and  LN,  _L  to  KL,  form- 
ing the  rectangular  parallelepiped  T. 

Consider  FI  the  base  of  S,  and  EF  its  altitude. 

Then  R=s  (574). 

Also  B  =  B'  (360). 

Consider  FP  the  base  of  S,  KM  the  base  of  T,  and  KL  its 
altitude. 

Then  s  =  T;  also  B'  ^  C  (574,  134). 

Hence  E  =  T  and  B  =  C  (Ax.  1). 

And  the  altitude  of  T  is  H  (509). 

But  the  volume  of  T  =  C  •  H  (582). 

.-.  Vof^  =  B-H  (Ax.  6).  Q.E.D. 


328  BOOK  VII.     SOLID   GEOMETRY 

585.  COROLLARY.    Two   parallelepipeds   having   equal  alti- 
tudes and  equal  bases  are  equal.  (Ax.  i.) 

586.  COROLLARY.     Two  parallelepipeds  having  equal  altitudes 
are  to  each  other  as  their  bases. 

Proof :  Q  =  B  •  n  and  E  =  Br  -  H.      .•.§  =  —    (584,  Ax.  3). 

R     it' 

587.  COROLLARY.     Two  parallelepipeds  having  equal  bases 
are  to  each  other  as  their  altitudes.  (?.) 

588.  COROLLARY.     Any    two    parallelepipeds    are    to    each 
other  as  the  products  of  their  bases  by  their  altitudes.         (?.) 

PROPOSITION  XII.     THEOREM 

589.  The  volume  of  a  triangular  prism  is  equal  to  the 
product  of  its  base  by  its  altitude. 


Given  :  Triangular  prism  ACD-F  \  base  =  B ;  alt.  =  H. 
To  Prove :  Volume  of  ACD-F  =  B  -  n. 

Proof:  Construct  parallelepiped  AS  having  as  three  of  its 
lateral  edges  AE,  CF,  DG. 

Vol.  AS  =  AGED  -  H  (584). 

Hence          J  volume  of  AS  =  \  AGED  -  H  (Ax.  3). 

But         \  volume  of  AS=  volume  of  prism  ACD-F     (575). 

And  \  ACRD=  B  (126). 

.-.  volume  of  ACD-F  =  B  •  H  (Ax.  6).     Q.E.D. 


PRISMS  329 

PROPOSITION  XIII.     THEOREM 

590.   The  volume  of  any  prism  is  equal  to  the  product  of  its 
base  by  its  altitude. 


Given  :  Prism  AD ;  base  =  to  B  ;  altitude  =  to  H. 
To  Prove  :  Vol.  of  AD  =  B  •  H. 

Proof:  Through  any  lateral  edge,  AC,  and  other  lateral 
edges  not  adjoining  AC,  pass  planes  cutting  the  prism  into 
triangular  prisms  I,  II,  III,  having  bases  R,  8,  T,  respectively. 

Vol.  of  prism      I  =  R  •  H 

'   (589). 


Vol.  of  prism    II  =  s  •  H 
Vol.  of  prism  III  =  T  •  H 


Adding, 


Vol.  of  prism  AD   =  (#  -f  s  +  r)  //  =  B  •  H    (Ax.  2).   Q.E.D. 

591.  COROLLARY.    Two  prisms  having  equal  altitudes  and 
equal  bases  are  equal. 

592.  COROLLARY.    Two  prisms  having  equal  altitudes  are  to 
each  other  as  their  bases. 

593.  COROLLARY.     Two  prisms  having  equal  bases   are  to 
each  other  as  their  altitudes. 

594.  COROLLARY.     Any  two  prisms  are  to  each  other  as  the 
products  of  their  bases  by  their  altitudes. 


330  BOOK   VII.     SOLID   GEOMETRY 

ORIGINAL  EXERCISES 

1.  Which  rectangular  parallelepiped  contains  the  greater  volume,  one 
whose  edges  are  5  in.,  7  in.,  0  in.,  or  one  whose  edges  are  4  in.,  6  in.,  13  in.? 

2.  The  base  of  a  prism  is  a  right  triangle  whose  legs  are  8  m.  and 
12  m.  and  the  altitude  of  the  prism  is  20  m.     Find  its  volume. 

3.  During  a  rain,  half  an  inch  of  water  fell.     How  many  gallons  fell 
on  a  level  ten-acre  park,  allowing  7£  gal.  to  the  cubic  foot  ? 

4.  Counting  38  cu.  ft.  of  coal  to  a  ton,  how  many  tons  will  a  coal  bin 
18  ft.  long,  6  ft.  wide,  and  9|  ft.  deep  contain,  when  even  full? 

6.   How  many  faces  has  a  parallelepiped?  edges?  vertices?     How 
many  faces  has  a  hexagonal  prism?  edges?  vertices? 

6.  Every  lateral  face  of  a  prism  is  parallel  to  the  lateral  edges  not 
in  that  face. 

7.  Every  lateral  edge  of  a  prism  is  parallel  to  the  faces  that  do  not 
contain  it. 

8.  Every  plane  containing  one  and  only  one  lateral  edge  of  a  prism 
is  parallel  to  all  the  other  lateral  edges. 

9.  Any  lateral  face  of  a  prism  is  less  than  the  sum  of  the  other  lateral 
faces. 

10.  The  diagonals  of  a  rectangular  parallelepiped  are  equal. 
Proof:  Pass  the  plane  ACGE.     This  is  a  rectangle  (?),  etc. 

11.  The  four  diagonals  of  a  parallelepiped  bisect 
each  other. 

[First  prove  that  one  pair  bisect  each  other;  thus 
prove  that  any  pair  bisect  each  other,  etc.] 

12.  Two  triangular  prisms  are  equal  if  their  lateral 
faces  are  equal  each  to  each. 

13.  Any  prism  is  equal  to  the  parallelepiped  hav- 
ing the  same  altitude  and  an  equal  base. 

14.  The  square  of  the  diagonal  of  a  rectangular  parallelepiped  is  equal 
to  the  sum  of  the  squares  of  its  three  dimensions. 

To  Prove  :  AC*  =  AE*  +  ED*  +  DC*. 

Proof:  AD  is  the  hypotenuse  of  rt.  &AED,  and 
AC,  of  rt.  A  A  CD. 

16.  The  diagonal  of  a  cube  is  equal  to  the  edge 
multiplied  by  V3.  A  E 


ORIGINAL   EXERCISES  331 

16.  The  volume  of  a  triangular  prism  is  equal  to  half  the  product  of 
the  area  of  any  lateral  face  by  the  perpendicular  drawn  to  that  face 
from  any  point  in  the  opposite  edge. 

17.  Every   section  of  a  prism  made  by  a  plane 
parallel  to  a  lateral  edge  is  a  parallelogram. 

To  Prove :  LMRN  a  O.     Proof:  LM  is  II  to  NR 
(?).    LN  and  MR  are  each  II  to  any  edge.    (Explain.) 

18.  Every  polyhedron  has  an  even  number  of  face 

angles. 

Proof:  Consider  the  faces  as  separate  polygons.  The  number  of  sides 
of  these  polygons  =  double  the  number  of  edges  of  the  polyhedron. 
(Explain.)  But  the  number  of  sides  of  these  polygons  =  the  number  of 
their  angles,  that  is,  the  number  of  face  angles.  .*.  the  number  of  face 
angles  =  double  the  number  of  edges  =  an  even  number  (V). 

19.  There  is  no  polyhedron  having  fewer  than  6  edges. 

20.  Find  the  contents  and  total  area  of  a  room  7  m.  x  5  m.  x  3  in. 

21.  Find  the  volume,  lateral  area,  and  total  area  of  an  8-inch  cube. 

22.  A  right  prism  whose  height  is  12  ft.  has  for  its  base  a  right  tri- 
angle whose  legs  are  6  ft.  and  8  ft.     Find  the  volume,  lateral  area,  and 
total  area  of  the  prism. 

23.  Find  the  altitude  of  a  rectangular  parallelepiped  whose  base  is 
21  in.  x  30  in.,  equivalent  to  a  rectangular  parallelepiped  whose  dimen- 
sions are  27  in.  x  28  in.  x  35  in. 

24.  A  cube  and  a  rectangular  parallelepiped  whose  edges  are  6  in., 
16  in.,  and  18  in.,  have  the  same  volumes.     Find  the  edge  of  the  cube. 

25.  Find  the  volume  of  a  rectangular  parallelepiped  whose  total  area 
is  620  sq.  in.  and  whose  base  is  14  in.  x  9  in. 

26.  How  many  bricks  each  8  in.  x  2|  in.  x  2  in.  will  be  required  to 
build  a  wall  22  ft.  x  3  ft.  x  2  ft.  (not  allowing  for  mortar)  ? 

27.  If  a  triangular  prism  is  20  in.  high  and  each  side  of  its  base  is 
8  in.,  how  many  cubic  inches  does  it  contain  ? 

28.  Find  the  lateral  area,  total  area,  and  volume  of  a  regular  hexag- 
onal prism  each  side  of  whose  base  is  10  in.  and  whose  altitude  is  15  in. 

29.  A  box  is  12  in.  x  9  in.  x  8  in.     What  is  the  length  of  its  diagonal  ? 

30.  Each  edge  of  a  cube  is  8  in.     Find  its  diagonal. 

31.  The  diagonal  of  a  cube  is  10  V3  in.     Find  its  edge,  volume,  and 
total  area. 


332  BOOK   VII.     SOLID   GEOMETRY 

32.  A  trench  is  180  ft.  long  and  12  ft.  deep,  7  ft.  wide  at  the  top  and 
4  ft.  at  the  bottom.    How  many  cubic  yards  of  earth  have  been  removed? 

33.  A  metallic  tank,  open  at  the  top,  is  made  of  iron  2  in.  thick;  the 
internal  dimensions  of  the  tank  are,  4  ft.  8  in.  long,  3  ft.  6  in.  wide,  4  ft. 
4  in.  deep.     Find  the  weight  of  the  tank  if  empty;  if  full  of  water. 

[Water  weighs  62^  Ib.  to  the  cubic  foot  and  iron  is  7.2  times  as  heavy 
as  water.] 

34.  The  base  of  a  right  parallelepiped  is  a  rhombus  whose  sides  are 
each  25  in.,  and  the  shorter  diagonal  is  14  in.    The  height  of  the  parallel- 
epiped is  40  in.     Find  its  volume  and  total  surface. 

35.  If  the  diagonal  of  a  cube  is  12  ft.,  find  its  surface. 

36.  If  the  total  surface  of  a  cube  is  54  sq.  ft.,  find  its  volume. 

37.  A  right  prism  whose  altitude  is  25  in.  has  for  its  base  a  triangle 
whose  sides  are  11  in.,  13  in.,  20  in.     Find  its  lateral  area,  total  area,  and 
volume. 

PYRAMIDS 

595.  A  pyramid  is  a  polyhedron,  one  of  whose  faces  is  a 
polygon  and  whose  other   faces  are  all   triangles  having  a 
common  vertex. 

The  lateral  faces  of  a  pyramid  are  the  triangles. 

The  lateral  edges  of  a  pyramid  are  the  intersections  of  the 
lateral  faces.  The  vertex  of  a  pyramid  is  the  common  vertex 
of  all  the  lateral  faces. 

The  base  of  a  pyramid  is  the  face  opposite  the  vertex. 

The  lateral  area  of  a  pyramid  is  the  sum  of  the  areas  of 
the  lateral  faces.  The  total  area  of  a  pyramid  is  the  sum  of 
the  lateral  area  and  the  area  of  the  base. 

The  altitude  of  a  pyramid  is  the  perpendicular  distance 
from  the  vertex  to  the  plane  of  the  base. 

A  triangular  pyramid  is  a  pyramid  whose  base  is  a  triangle. 
It  is  called  also  a  tetrahedron.  (See  551.) 

596.  A  regular  pyramid  is  a  pyramid  whose  base  is  a  regu- 
lar polygon  and  whose  altitude,  from  the  vertex,  meets  the 
base  at  its  center. 


PYRAMIDS  333 


PYRAMIDS  RKGULAR  TRUNCATKD       FRUSTUM  OP 

PYRAMIDS  PYRAMID          A  PVRAMID 

The  slant  height  of  a  regular  pyramid  is  the  line  drawn  in 
a  lateral  face,  from  the  vertex  perpendicular  to  the  base  of 
the  triangular  face.  It  is  the  altitude  of  any  lateral  face. 

597.  The  frustum  of  a  pyramid  is  the  part  of  a  pyramid 
included  between  the  base  and  a  plane  parallel  to  the  base. 

The  altitude  of  a  frustum  of  a  pyramid  is  the  perpendicular 
distance  between  the  planes  of  its  bases. 

The  slant  height  of  the  frustum  of  a  regular  pyramid  is 
the  perpendicular  distance,  in  a  face,  between  the  bases  of 
that  face. 

A  truncated  pyramid  is  the  part  of  a  pyramid  included 
between  the  base  and  a  plane  cutting  all  the  lateral  edges. 

PRELIMINARY  THEOREMS 

598.  THEOREM.     The  lateral  edges  of  a  regular  pyramid  are 
all  equal.  (504,  II.) 

599.  THEOREM.     The  lateral  faces  of  a  regular  pyramid  are 
congruent  isosceles  triangles.  (598,  78.) 

600.  THEOREM.     The  lateral  edges  of  the  frustum  of  a  regu- 
lar pyramid  are  all  equal.  (Ax.  2.) 

601.  THEOREM.     The  lateral  faces  of  the  frustum  of  a  regu- 
lar pyramid  are  congruent  isosceles  trapezoids.  (484.) 

602.  THEOREM.     The  lateral  faces   of  the  frustum   of  any 
pyramid  are  trapezoids.  (?.) 

603.  THEOREM.     The  slant  height  of  a  regular  pyramid  is 
the  same  length  in  all  the  lateral  faces. 


334  BOOK    VII.     SOLID   GEOMETRY 

THEOREMS   AND   DEMONSTRATIONS 

PROPOSITION  XIV.     THEOREM 

604.  The  lateral  area  of  a  regular  pyramid  is  equal  to 
half  the  product  of  the  perimeter  of  the  base  by  the  slant 
height. 

0, 


Given :   Regular  pyramid  O-ABCDE ;  lateral  area  =  to  L  ; 
perimeter  of  base  =  to  P;  slant  height  OH=  to  s. 

To  Prove  :  L  =  \  P  •  s. 

Proof :  Area  A  AOB  =  ±  AB  •  s  j 
Area  A  BOC  =  ^  BC  •  s\ 
etc.  etc. 


Adding,    Lateral  area  =  |  AB  •  s  +  |  BC  -  s  +  etc.    (Ax.  2). 
That  is,  L  =  \(AB  +  £C  +  etc.)  .  s,  or, 

Lateral  area,  L  =  \  P  •  s  (Ax.  6).     Q.E.D. 


Ex.  1.  Prove  that  the  bases  of  any  frustum  of  a  pyramid  are  mutually 
equiangular. 

Ex.  2.  The  foot  of  the  altitude  of  a  regular  pyramid  drawn  from  the 
vertex,  coincides  with  the  center  of  the  circles  inscribed  in,  and  circum- 
scribed about,  the  base. 

Ex.  3.  The  sum  of  the  medians  of  the  lateral  faces  of  the  frustum  of 
a  pyramid  is  equal  to  half  the  sum  of  the  perimeters  of  the  bases. 

Ex.  4.  To  what  rectangle  is  the  lateral  area  of  a  regular  pyramid 
equal?  the  total  area? 


PYRAMIDS 
PROPOSITION  XV.     THEOREM 


335 


605.  The  lateral  area  of  the  frustum  of  a  regular  pyramid  is 
equal  to  half  the  sum  of  the  perimeters  of  the  bases  multiplied 
by  the  slant  height. 


Given:   (?). 

To  Prove  :  L  =  |(p  +^)  •  8- 

Proof  :  Area  trapezoid  C  I  =  J(CD  +  HI)  •  s 
Area  trapezoid  BH  =  \(BC  +  GH)  •  s 
Area  trapezoid  AG  =  \(AB  +  FG)  •  8 

Adding,      Lateral  area,  L  —  |(J»  +  p)  .  s 


(?), 
(?). 

(?)     etc. 

Explain. 
Q.E.D. 


•  Ex.  1.  Using  L  for  lateral  area  and  B  for  area  of  base  express  the 
formula  for  the  total  area,  T,  of  a  regular  pyramid. 

Ex.  2.  The  slant  height  of  a  regular  pyramid  whose  base  is  a  square, 
of  which  each  side  is  8  ft.,  is  15  ft.  Find  the  lateral  area;  the  total 
area. 

Ex.  3.  A  regular  pyramid  stands  on  a  hexagonal  base  16  in.  on  a 
side,  and  the  slant  height  is  2  ft.  Find  the  lateral  and  total  areas. 

Ex.  4.  The  slant  height  of  the  frustum  of  a  regular  pyramid  is  12  in., 
and  the  bases  are  squares,  10  in.  and  6  in.  on  a  side,  respectively.  Find 
the  lateral  area  and  the  total  area. 

Ex.  6.  How  many  square  feet  of  tin  will  be  required  to  line  a  vat  in 
the  form  of  the  frustum  of  a  regular  pyramid,  having  inside  measure- 
ments as  follows  :  the  slant  height  is  14  ft.  ;  the  bases  are  regular  hexa- 
gons whose  sides  are  7  ft.  and  6  ft.  respectively. 

Ex.  6.  The  lateral  edge  of  a  pyramidal  church  spire  is  61  ft.  Each 
side  of  its  octagonal  base  is  22  ft.  What  will  be  the  cost  of  painting  the 
spire  at  2^  a  square  foot  ? 


336 


BOOK   VII.     SOLID   GEOMETRY 


PROPOSITION  XVI.     THEOREM 

606.   If  a  pyramid  is  cut  by  a  plane  parallel  to  the  base : 
I.  The  lateral  edges  and  altitude  are  divided  proportionally. 
n.  The  section  is  a  polygon  similar  to  the  base. 

III.  The  area  of  the  section  is  to  the  area  of  the  base  as  the 
square  of  its  distance  from  the  vertex  is  to  the  square  of  the 
altitude  of  the  pyramid. 


Given  :  Pyr.  O-ABCDE-,    plane  Fi 

OL. 


to  the  base  ;  altitude 


To  Prove 


II. 
III. 


OF_=OGs=OH==        =03f 
OA       OB       OC  OL 

Section  FI  is  similar  to  the  base  AD. 

section  FI  _  OM2 
base  AD        QL2' 

Proof:  I.    Imagine  a  plane  through  o  II  to  plane  AD. 
This  plane  is  J_  to  OL 
And  II  to  plane  FI 

OF  _  OG  _  OH  _         _  OM 
'  OA~~  OB~  OC~         7  OL 

II.  FG  is  II  to  AB,  GH  is  II  to  BC,  etc. 

.-.  Z  FGH=  Z  ABC  ;    Z  GUI—  Z  BCD;   etc. 
That  is,  the  polygons  are  mutually  equiangular. 
A  OFG  is  similar  to  A  OAB\  A  OGH,  to  A  OBC;  etc.    (305). 


(496). 
(489). 


(484). 
(499). 


PYRAMIDS  337 

...  ™  =(««)=  ^  =  (25?)=  *U etc.  (313). 

AH      \OBJ      BC       \OCj      CD 
.'.  section  FI  is  similar  to  base  AD     (301).     Q.E.D. 

III.       Section  FI  is  similar  to  base  AD  (II,  above), 

section  FI      1G2  ^OT*\ 

— : —  — -•  (OlO). 

base  AD       2^2 
Now  A  OFG  is  similar  to  A  OAB  (305). 

.-.™  =  ™:  (313). 

AB       OA 
r>    4  OF        OM  ,T       T          N 

But  -=  -  (I,  above). 

OA       OL 

FG       OM 

.'.  --  =  -  (Ax.  1). 

AB        OL 

And  ^  =  ^  (287). 

AB"       OL 

Hence  section  f/=o^ 

K—  ^s       5Z2 


Ex.  1.  The  bases  of  the  frustum  of  a  regular  pyramid  are  equilateral 
triangles  whose  sides  are  12  in.  and  20  in.,  respectively.  The  slant 
height  is  40  in.  Find  the  lateral  area  ;  the  total  area. 

Ex.  2.  The  bases  of  the  frustum  of  a  regular  pyramid  are  regular  hexa- 
gons whose  sides  are  8  in.  and  18  in.,  respectively.  The  slant  height  is 
25  in.  Find  the  lateral  area  and  total  area. 

Ex.  3.  A  pyramid  whose  altitude  is  10  in.  and  whose  base  contains 
80  sq.  in.  is  cut  by  a  plane  bisecting  the  altitude  and  parallel  to  the  base. 
Find  the  area  of  the  section  of  the  pyramid  made  by  this  plane. 

Ex.  4.  Cutting  a  pyramid  whose  altitude  is  16  ft.  is  a  plane  parallel 
to  the  base  and  6  ft.  from  it.  The  area  of  the  base  is  192  sq.  ft.  What 
is  the  area  of  the  section  ? 

Ex.  5.  If  a  plane  parallel  to  the  base  of  a  pyramid  bisects  the  altitude, 
how  does  the  area  of  the  section  compare  with  the  area  of  the  base  ? 

Ex.  6.  Two  planes,  parallel  to  the  base  of  a  pyramid,  trisect  the  alti- 
tude. How  do  the  areas  of  the  sections  compare  with  the  area  of  the  base  ? 


338  BOOK  VII.     SOLID   GEOMETRY 

PROPOSITION  XVII.     THEOREM 

607.  If  two  pyramids  have  equal  altitudes  and  equal  bases, 
sections  made  by  planes  parallel  to  the  bases  and  at  equal 
distances  from  the  vertices  are  equal. 


Given:  Pyramids  O-A BCD E  and  o'-PQRS-,  alt.  OL  =  o' Lf ; 
base  AD  =  base  PR  ;  planes  of  the  sections  FI  and  TV  II  to  the 
bases;  OM  =  O'M'. 


To  Prove :  Section  FI  =  section  TV. 

:2 


Proof: 
But 


section  FI 


OM"       T  section  TV 
and 


O'M 


base  AD        QL2  base  PR 

OM=0'M'  and  OL  =  Or  L1 

section  FI  _  section  TV 
base  AD          base  PR 

Now  base  AD  =  base  PR 

Multiplying,         section  FI=  section  TV 


5 (806,  III). 

(Hyp-)- 
(Ax.  1). 

(Hyp.). 
(Ax.  3). 

Q.E.D. 


608.  If  a  plane  is  passed  parallel  to  the  base  of  a  pyramid,  intersect- 
ing all  the  lateral  edges,  and  upon  this  section,  as  a  base,  a  prism  is  con- 
structed wholly  inside  the  pyramid,  but  having  one  lateral  edge  in  a 
lateral  edge  of  the  pyramid,  this  prism  is  called  an  inscribed  prism. 

If  upon  this  section  as  a  base  a  prism  is  constructed  partly  outside  the 
pyramid,  having  one  lateral  edge  in  one  of  the  lateral  edges  of  the  pyra- 
mid, this  prism  is  called  a  circumscribed  prism. 


PYRAMIDS  339 

PROPOSITION  XVIII.     THEOREM 

609.  The  volume  of  a  triangular  pyramid  is  the  limit  of  the 
sum  of  the  volumes  of  a  series  of  inscribed  or  circumscribed 
prisms,  having  equal  altitudes,  if  the  number  of  prisms  is  in- 
definitely increased. 


Given :  Triangular  pyramid  O-ABC,  having  a  series  of 
prisms  inscribed  in  it,  and  another  series  circumscribed 
about  it,  all  the  prisms  having  equal  altitudes. 

To  Prove  :  O-ABC  is  the  limit  of  the  sum  of  each  series  of 
prisms  as  their  number  is  indefinitely  increased. 

Proof:  Denote  the  volume  of  the  pyramid  by  F,  the  sum  of 
the  volumes  of  the  series  of  inscribed  prisms  by  8&  and  the  sum 
of  the  volumes  of  the  series  of  circumscribed  prisms  by  Sc. 

The  uppermost  circumscribed  prism  =  the  uppermost  in- 
scribed prism  (591). 

The  second  pair  of  prisms  also  are  equal  (?). 

And  so  on,  until  the  last  circumscribed  prism,  D-ABC,  re- 
mains, for  which  there  is  no  equivalent  inscribed  prism. 

Hence  it  is  evident  that  8C  —  8t  =  D-ABC,  the  lowest 
circumscribed  prism. 

Now  by  indefinitely  increasing  the  number  of  the  prisms, 
the  altitude  of  D-ABC  becomes  indefinitely  small,  and  hence 
the  volume  of  D-ABC  approaches  zero  as  a  limit. 

The  altitude  can  never  actually  equal  zero,  nor  can  the 
volume  equal  zero.  Hence  Sc  —  S{  can  be  made  less  than 
any  mentionable  quantity,  but  cannot  equal  zero. 

ROBBINS'S   NEW   SOLID   GEOM. — 7 


340  BOOK   VII.     SOLID   GEOMETRY 


Now  sc  =  sc 

and         F  >  8t          (Ax.  5). 
.*.  Sc  —  V  <  Sc  —  Si  (Ax.  9). 


v  <  sc          (Ax.  5). 


and  Si  = 


.'.  V-8t  <  Se-8{  (Ax.  7). 


That  is,  8e  —  V  and  F  —  St  are  each  less  than  Sc  —  8#  which 
itself  approaches  zero. 

Hence  8e  —  F  approaches  zero  and  F  —  S{  approaches  zero. 

.-.  sc  approaches  F  as  a  limit,  and  S{  approaches  F  as  a 
limit  (227).  (See  note  on  p.  xviii.) 

Q.E.D. 

Ex.  1.  If  a  plane  is  passed  parallel  to  the  base  of  a  pyramid,  cutting 
the  lateral  edges,  the  section  is  to  the  base  as  the  square  of  the  lateral 
edge  of  the  pyramid  cut  away  by  this  plane  is  to  the  square  of  the  lateral 
edge  of  the  original  pyramid.  [See  proof  of  60C,  III.] 

Ex.  2.  If  two  pyramids  have  equal  altitudes  and  are  cut  by  planes 
parallel  to  the  bases  and  at  equal  distances  from  the  vertices,  the  sec- 
tions formed  will  be  to  each  other  as  the  bases  of  the  pyramids. 

Ex.  3.  In  the  figure  of  609  prove  that  the  planes  of  the  faces  of  the 
prisms  that  are  opposite  OC  are  parallel  to  OC  and  AB. 

Ex.  4.     State  the  theorems  leading  up  to  the  theorem  of  575. 

Ex.  5.     State  the  theorems  leading  up  to  the  theorem  of  584. 

Ex.  6.     State  the  theorems  leading  up  to  the  theorem  of  590. 

Ex.  7.  The  base  of  a  pyramid  is  180  sq.  in.  and  its  altitude  is  15  in. 
What  is  the  area  of  the  section  made  by  a  plane  parallel  to  the  base,  and 
5  in.  from  the  vertex? 

Ex.  8.  The  base  of  a  pyramid  is  200  eq.  in.,  and  its  altitude  is  12  in. 
At  what  distance  from  the  vertex  must  a  plane  be  passed  so  that  the 
section  shall  contain  half  the  area  of  the  base  ? 

Ex.  9.  The  base  of  a  pyramid  is  B  sq.  in.,  and  the  altitude  is  h  in. 
How  far  from  the  vertex  must  a  plane  parallel  to  the  base  be  passed  so 
that  the  area  of  the  section  shall  contain  I  B  sq.  in.  ?  \  B  sq.  in.  ?  Find 
the  area  of  the  section  if  the  plane  is  passed  \  h  in.  from  the  vertex ;  if  it 
is  passed  \  h  in.  from  the  vertex. 

Ex.  10.  A  pyramid  having  an  altitude  of  2  ft.  and  a  base  which  is  an 
equilateral  triangle  of  8  in.  on  a  side,  is  cut  by  a  plane  parallel  to  the  base 
and  18  in.  from  it.  Find  the  area  of  the  section. 


PYRAMIDS 


341 


PROPOSITION  XIX.     THEOREM 

610.   Two  triangular  pyramids  having  equal  altitudes   and 
equal  bases  are  equal. 


Given:  Triangular  pyramids  O-ABC  and  of-AfB'c(  having 
equal  altitudes  and  base  ABC  =  base  AfBfCf. 

To  Prove :  O-ABC  =  O'-A'B'C'. 

Proof :   Divide    the    altitude    of    each    pyramid   into   any 
number  of  equal  parts. 

Through  these  points  of  division  pass  planes  II  to  the  bases, 
forming  triangular  sections. 

Upon  these  sections  as  bases  construct  inscribed  prisms. 
Denote  the  volumes  of  the  pyramids  by  v  and  V1 ',  and  the 
sums  of  the  volumes  of  these  series  of  prisms  by  S  and  8f. 

The  corresponding  sections  are  equal        (607). 
.*.  the  corresponding  prisms  are  equal       (591). 
Hence  8  =  sf  (Ax.  2). 

By  indefinitely  increasing  the  number  of  equal  parts  into 
which  the  altitudes  are  divided,  the  number  of  prisms  be- 
comes indefinitely  great.     .-.  S  approaches  F  as  a  limit  (609). 
And  s'  approaches  V1  as  a  limit  (?). 

.-.  F=  V'  (229). 

That  is,  O-ABC  =  O'-A'B'C'.  Q.E.D. 


NOTE.  As  in  plane  geometry,  A  ABC  is  the  same  as  ABAC,  so  in 
solid  geometry  the  pyramid  O-ABC  is  the  same  as  the  pyramid  A-BCO 
or  B-ACOor  C-ABO. 


342 


BOOK    VII.     SOLID   GEOMETRY 


PROPOSITION  XX.     THEOREM 

611.  The  volume  of  a  triangular  pyramid  is  equal  to  one 
third  the  product  of  its  base  by  its  altitude. 


Given :  Triangular  pyramid  O-AMC,  whose  base  =  B  and 
altitude  =  H. 

To  Prove :  Volume  O-AMC  =  £  B  •  H. 

Proof :  Construct  a  prism  AMC-DOE,  having  AMC  as  its 
base,  and  OM  as  one  of  its  lateral  edges.  Pass  a  plane  through 
DO  and  o<7,  cutting  the  face  AE  in  line  CD.  The  prism  is  now 
divided  into  three  triangular  pyramids. 

In  pyramids  O-AMC  and  C-ODE,  the  altitudes  are  =  (509). 

The  bases  AMC  and  ODE  are  ^  (553). 

.-.  pyramid  O-AMC  =  pyramid  C-ODE  (610). 

In  pyramids  C-AMO  and  C-AOD,  the  altitudes  are  the  same 

line,  u  _L  from  C  to  plane  DM  (491). 

The  bases  AMO  and  AOD  are  ^  (126). 

.*.  pyramid  C-AMO  =  pyramid  C-AOD         (610). 

Hence  O-AMC  =  C-ODE  =  C-AOD  (Ax.  1). 

That  is,  O-AMC  =  J  the  prism. 

But  the  volume  of  the  prism   =  B  •  H  (?). 

.-.  volume  of  pyramid  O-AMC  =£/*•//     (Ax.  6).     Q.E.D. 


Ex.  1.     In  the  figure  of  611  prove  pyramid  0-ACD  =  O-CDE. 

Ex.  2.  The  area  of  the  base  of  a  triangular  pyramid  is  30  sq.  in.,  and 
its  altitude  is  20  in.  Find  the  volume.  Find  the  volume  of  the  prism 
having  the  same  base  and  altitude. 


PYRAMIDS  343 

PROPOSITION  XXI.     THEOREM 

612.  The  volume  of  any  pyramid  is  equal  to  one  third  the 
product  of  its  base  by  its  altitude. 

O, 


Given :  Pyramid  o-CDEFG,  whose  base  =  B  and  altitude 
=  H. 

To  Prove :  Volume  of  o-CDEFG  =  %  n  •  H. 

Proof :  Through  any  lateral  edge,  OC,  and  lateral  edges 
not  adjoining  OC,  pass  planes  dividing  the  pyramid  into  tri- 
angular pyramids. 

Vol.  of  O-CFG      =  1  CFG  -  H 

Vol.  of  0-CDE       =  1  CDE  •  H 

Vol.  of  0-CEF       =  1  CEF-  H 


(611). 
Adding, 
Vol.  of  O-CDEFG  =  J B  •  H      (Ax.  2,  4).  Q.E.D. 

613.  COROLLARY.     Any  two  pyramids  having  equal  altitudes 
and  equal  bases  are  equal.  (Ax.  1.) 

614.  COROLLARY.     Two  pyramids  having  equal  altitudes  are 
to  each  other  as  their  bases.  (Prove.) 

615.  COROLLARY.     Two  pyramids  having  equal  bases  are  to 
each  other  as  their  altitudes.  (Prove.) 

616.  COROLLARY.     Any  two  pyramids  are  to  each  other  as 
the  products  of  their  bases  by  their  altitudes.  (Prove.) 

Ex.  The  altitude  of  a  pyramid  is  15  ft.,  and  each  side  of  its  square 
base  is  8  ft.  Find  the  volume.  What  is  the  volume  of  a  prism  having 
the  same  base  and  altitude  ? 


344  BOOK   VII.     SOLID   GEOMETRY 

PROPOSITION  XXII.     THEOREM 

617.  The  volume  of  the  frustum  of  a  triangular  pyramid  is 
equal  to  one  third  the  altitude  multiplied  by  the  sum  of  the 
lower  base,  the  upper  base  and  a  mean  proportional  between 
the  bases  of  the  frustum. 


Given :  The  frustum  ED  of  a  triangular  pyramid  whose 
lower  base  =  B  ;  upper  base  =  b  ;  altitude  =  H. 

To  Prove :  Volume  of  ED  =  J  H  IB  +  b  +  V#  •  b~\. 

Proof :  Pass  a  plane  through  edge  CE  and  vertex  s,  and 
another  through  edge  RS  and  vertex  E,  dividing  the  frustum 
into  three  triangular  pyramids,  S-CDE,  E-RST,  E-CRS. 

I.  S-CDE  =  I  H-.B 

II.  E-RST  =  %H  •  b 

III.    We  shall  now  prove 
E-CRS = 
E-CSD      A 
.E-CflS " 
A  C&D^ 
A  CRS 
E-CSD 
E-CRS ' 


(368). 
(Ax.  1). 


Likewise  ^  =  ^  (?). 

8-ERT      AERT 


PYRAMIDS  345 

And 


But 


Hence 

That  is,  

E-CRS       £ 

(Substituting  from  I  and  II). 

.-.  E-CRS  =  V±  n  •  B  •  J  H  -  b  =  i  //  •  VB  •  b  (289). 

.-.  Volume  of  the  frustum  =  £#[5  + 6+ Vfl.fr]     (Ax.  2). 

Q.E.D. 

NOTE.     The  theorem  of  617  is  sometimes  stated  thus: 
The  frustum  of  a  triangular  pyramid  is  equal  to  the  sum  of  three 
pyramids  whose  altitudes  are  the  same  as  the  altitude  of  the  frustum  and 
whose  bases  are  the  lower  base,  the  upper  base,  and  a  mean  proportional 
between  the  bases  of  the  frustum. 


A  CER       CE 

(?)• 
(?). 

(606,  II). 
.(?)• 
(Ax.  1). 

A  ERT      RT 

S-ERT      RT 
CDE  and  RST  are  similar 
CD_CE 
RS~  RT 
E-CSD      S-CER  or  E-CRS 

E-CRS                 S-ERT 
J  H  •  B  _  E-CRS 

E-CRS       I  II  •  I 

Ex.  1.  Find  the  volume  of  the  pyramid  whose  altitude  is  18  in.  and 
whose  base  is  10  in.  square. 

Ex.  2.  Find  the  volume  of  the  frustum  of  a  triangular  pyramid  whose 
altitude  is  20  in.  and  the  areas  of  whose  bases  are  18  sq.  in.  and  32  sq.  in. 

Ex.  3.  If,  in  the  formula  for  the  volume  of  the  frustum  of  a  triangu- 
lar pyramid,  b  =  B,  show  that  the  formula  becomes  the  correct  formula 
for  the  volume  of  a  prism.  And,  again,  if  b.=  0,  show  that  the  formula 
of  617  becomes  the  correct  formula  for  the  volume  of  a  pyramid. 

Ex.  4.  What  is  the  locus  of  the  vertices  of  all  pyramids  upon  the  same 
base  and  having  the  same  volume? 


Historical  Note.  The  Greeks,  as  early  as  the  fourth  century  B.C.,  knew 
that  the  pyramid  is  the  third  part  of  the  prism  having  the  same  base  and 
altitude.  Eudoxus,  an  Athenian  mathematician,  is  given  credit  for  the 
discovery  and  proof  of  this  truth. 


346  BOOK   VII.     SOLID   GEOMETRY 

PROPOSITION  XXIII.     THEOREM 

618.  The  volume  of  the  frustum  of  any  pyramid  is  equal  to 
one  third  the  altitude  multiplied  by  the  sum  of  the  lower  base, 
the  upper  base,  and  a  mean  proportional  between  the  bases  of 
the  frustum.  o  P 


Given :  Pyr.  O-ADEFG ;  frustum  Ar F,  whose  lower  base 
=  U,  upper  base  =  #,  altitude  =  H. 

To  Prove :  Volume  of  frustum  =  i  H  [B  +  b  +  VlTT| . 

Proof :  Construct  a  A  QRS  =  to  polygon  AF  (395). 

Upon  A  QRS  as  a  base,  construct  a  pyramid  whose  altitude 
=  the  altitude  of  O-ADEFG. 

Pass  a  plane  Q'R'S'  II  to  QRS  and  at  a  distance  from  QRS= to  H. 
Vol.  of  Q'R  =  %  H  [A  QRS  +  A  Q'R'S'  -f-  VA  QRS  -  A  Q'R'S'~\ 

(617). 

The  alt.  of  P-Q'R'S'  =  alt.  of  O-A'D'E'F'G'  (Ax.  2). 

Also  QRS  =  B  (Const.);  and  Q'R'S'  =  b  (607). 

.'.  vol.  of  O-ADEFG  =  vol.  of  P-QRS  (613). 

And  vol.  of  O-A'D'E'F'G'  =  vol.  of  P-Q'R'S'  (?). 

Subtracting,  vol.  of  frustum  A'F—  vol.  of  frustum  Q'R 

(Ax.  2). 
Vol.  of  frustum  A'F  =  ^  H[fl  +  6+VB-&]    (Ax.  6). 

Q.E.D. 

Ex.  The  bases  of  the  frustum  of  a  pyramid  are  regular  hexagons 
whose  sides  are  10  in.  and  6  in.,  respectively.  The  altitude  of  the  frus- 
tum is  2  ft.  Find  its  volume. 


PYRAMIDS  347 

PROPOSITION  XXIV.     THEOREM 

619.  A  truncated  triangular  prism  is  equal  to  three  triangu- 
lar pyramids  whose  bases  are  the  base  of  the  prism  and  whose 
vertices  are  the  three  vertices  of  the  face  opposite  the  base 
(the  inclined  section). 


HI 


Given:  The  truncated  triangular  prism  ABC-RST,  whose 
base  is  ABC  and  whoso  opposite  vertices  are  JR,  S,  T.  Let  it 
be  divided  by  the  planes  ACS,  ABT,  BCR. 

To  Prove  :  ABC-liST  =  R-ABC  +  S-ABC  -f-  T-ABC  (HI). 
Proof  :  In  Fig.  I,  S-ABC  is  obviously  one  of  these  pyramids. 
In  Fig.  II,  A-CST  =  A-BCT  (613). 

That  is,  A-CST=  T-ABC. 

In  Fig.  Ill,  T-ARS  =  T-ABR  (613). 

T-ABR  =  C-ABR  (613). 

.'.  T-ARS  =  R-ABC  (Ax.  1). 

Now         ABC-RST=  T-ARS  +  S-ABC  +  A-CST          (Ax.  4). 

Hence   ABC-RST  =  R-ABC  -f  S-ABC  +  T-ABC       (Ax.  6). 

Q.E.D. 


Ex.  There  are  approximately  1£  cu.  ft.  in  a  bushel.  Find  the  capacity, 
in  bushels,  of  a  grain  elevator,  30  ft.  high,  in  the  shape  of  the  frustum  of 
a  square  pyramid,  and  having  bases  24  ft.  square  and  16  ft.  square. 


348 


BOOK   VII.     SOLID   GEOMETRY 


620.  COROLLARY.     The  volume  of  a  trun- 
cated triangular  prism  is  equal  to  the  prod- 
uct of  the  base  by  one  third  the  sum  of  the 
three  altitudes  drawn  to  the  base  from  the 
three  vertices  opposite  the  base. 

621.  COROLLARY.     The  volume  of  a  trun- 
cated right  triangular  prism  is  equal  to  the 

product  of  the  base  by  one  third  the  sum  of  its   lateral 
edges. 

622.  COROLLARY.    The  volume  of  any  trun- 
cated triangular  prism  is  equal  to  the  product 
of  its  right  section  by  one  third  the  sum  of  its 
lateral  edges. 

Proof  :  The  right  section  divides  the  solid 
into  two  truncated  right  prisms. 

Hence  the  volume  =  the  right  section  x  ^ 
lateral  edges. 


the  sum  of  the 
(621.)     Q.E.D. 


Ex.  1.  A  pyramid  whose  volume  is  V  and  whose  altitude  is  h,  is 
bisected  by  a  plane  parallel  to  the  base.  Find  the  distance  of  this  plane 
from  the  vertex. 

Ex.  2.  The  altitude  of  a  square  pyramid  each  side  of  whose  base  is 
6  ft.,  is  10  ft.  Parallel  to  the  base  and  2  ft.  from  the  vertex  a  plane  is 
passed.  Find  the  area  of  the  section.  Find  the  volumes  of  the  two 
pyramids  concerned,  and  hence  find  the  volume  of  the  frustum. 

Ex.  3.  The  base  of  a  pyramid  is  a  rhombus  whose  diagonals  are  7  m. 
and  10  m.  Find  the  volume  if  the  altitude  is  15  m. 

Ex.  4.  The  areas  of  the  bases  of  the  frustum  of  a  pyramid  are 
3  sq.  ft.  and  27  sq.  ft.  The  volume  is  104  cu.  ft.  Find  the  altitude. 

Ex.  6.  State  what  distances  must  be  known  in  order  to  find  the 
volume  of  a  truncated  right  triangular  prism,  and  of  a  truncated  oblique 
triangular  prism.  Now  explain  how  to  use  these  lengths  to  find  the 
volume. 


PYRAMIDS  349 

PROPOSITION  XXV.     THEOREM 

623.  Two  triangular  pyramids  (tetrahedrons)  having  a  tri- 
hedral angle  of  one  equal  to  a  trihedral  angle  of  the  other  are 
to  each  other  as  the  products  of  the  three  edges  including  the 
equal  trihedral  angles. 


Given :  Triangular  pyramids  S-ABC,  S-PQR ;  having  the 
trill.  A  at  S  equal ;  and  their  volumes  V  and  v' . 

To  Prove:    Z  =  Ml^l^. 

V1       SP  •  SQ  •  SB 

Proof :  Place  the  pyramids  so  that  the  equal  trihedral  A 
coincide.  Draw  the  altitudes  AX  and'PF  and  the  projection 
SXY,  of  the  edge  PS,  in  plane  SQR. 


A  SAX  is  similar  to  A  SPY 
V       A  SBC  -AX      A  SBC      AX 

(304). 
(616). 

(313). 

Vf       ASQR-PY      A  SQR       PY 
A  SBC  _  SB  -  SC     (374X  and  AX  _  SA 

ASQR~SQ-SR                             PY      SP 

But 

Hence,  by  substituting  in  the  first  equation  above, 
V       SB  •  SC      SA       SA  •  SB  •  SC 


Vf       SQ  •  SR       SP       SP  •  SQ  •  SR 


(Ax.  6).  Q.E.D. 


Ex.  1.     To  what  is  the  ratio  of  A  SAB  to  A  SPQ  equal? 
Ex.  2.     Reduce   the   formula  of   623   if    SA  =  SB  =  SC   and   SP  = 
SQ  =  SR. 


350  BOOK   VII.    SOLID  GEOMETRY 

PROPOSITION  XXVI.     THEOREM 

624.  In  any  polyhedron  the  number  of  edges  increased  by 
two  is  equal  to  the  number  of  vertices  increased  by  the  num- 
ber of  faces. 


Given :  A  polyhedron  ;  E  =  number  of  edges  ;  F  =  num- 
ber of  faces  ;  F  =  number  of  vertices. 

To  Prove :    E  +  2  =  v  +  F. 

Proof:  Suppose  the  surface  of  the  polyhedron  is  put  to- 
gether, face  by  face. 

For  one  face,  E  =  V  (145).     (Begin  with  the  base.) 

By  attaching  an  adjoining  face,  the  number  of  edges  is 
one  greater  than  the  number  of  vertices. 

That  is,  for  two  faces,  E  =  V  +  1. 

For  three  faces,  E=V+2. 

For  four  faces,  E=V+S. 

For  five  faces,  E  =  V  +  4. 

For  n  faces,  E=V  +  (n—l). 

For  F-l  faces,  E=  F  +  (F-2). 

By  attaching  the  last  face,  neither  the  number  of  edges 
nor  the  number  of  vertices  is  increased. 

That  is,  for  F  faces,  E  =  V  +  F  —  2. 

.-.  for  the  complete  solid,  ^  +  2  =  V+F   (Ax.  2).    Q.E.D. 

625.  COROLLARY.  In  any  polyhedron  the  difference  between 
the  number  of  edges  and  the  number  of  faces  is  two  less  than 
the  number  of  vertices,  that  is,  E  -  F  =  V  -  2. 


POLYHEDRONS 


351 


PROPOSITION  XXVII.     THEOREM 

626.  The  sum  of  all  the  face  angles  of  any  polyhedron  is 
equal  to  4  right  angles  multiplied  by  two  less  than  the  number 
of  vertices,  that  is,  S  A  —  (  F  —  2)  4  rt.  A  =  (  V  —  2)  360°. 


i      J 


Given:  A  polyhedron  ;  E  =  number  of  edges;  F=  number 
of  faces  ;  F  =  number  of  vertices. 

To  Prove :  Sum  of  all  the  face  A  =  (F -  2)  4  rt.  A. 

Or,  8A=  (r-2)  360°. 

Proof :  If  the  faces  are  considered  as  separate  polygons,  it 
is  obvious  that  each  edge  is  a  side  of  two  polygons,  that  is,  the 
number  of  sides  of  the  several  faces  =  2  E. 

.*.  the  number  of  vertices  of  all  the  polygons  =  2  E    (145). 

Suppose  an  exterior  Z  formed  at  each  of  these  2  E  vertices. 

Then  the  sum  of  the  exterior  A  of  each  face  =4  rt.  A   (?). 

Sum  of  int.  and  ext.  A  at  each  vertex  =  2  rt.  A  (?). 

.  -.  the  int.  A  +  ext.  A  at  all  the  2  E  vertices  =  4  E  rt.  A 
But  the  ext.  A  of  all  the     F  faces       =  4  Frt.  A  (?). 

.  *.  the  int.  A  of  these  polygons  =  4  E  rt.  A  —  4  F  rt.  A  (?). 

=  (E  -  F)  •  4  rt.  A. 

But  E-F=V-2  (625). 

.-.   SA  =  (V-V)  4rt.  Zs  =  (F-2)360°       (Ax.  6).  Q.E.D. 

627.  Prismatoid.  A  prismatoid  is  a  polyhedron  two  of 
whose  faces,  called  bases,  are  polygons  in  parallel  planes,  and 


352 


BOOK   VII.     SOLID   GEOMETRY 


whose  other  faces  are  triangles,  trapezoids,  or  parallelo- 
grams. 

The  altitude  of  a  prismatoid  is  the  distance  between  the 
planes  of  the  bases. 

The  mid-section  of  a  prismatoid  is  the  section  made  by  a 
plane  parallel  to  the  bases  and  bisecting  the  altitude. 

A  prismoid  is  a  prismatoid  whose  lateral  faces  are  either 
trapezoids  or  parallelograms. 

PROPOSITION  XXVIII.     THEOREM 

628.  The  volume  of  a  prismatoid  is  equal  to  the  product 
of  one  sixth  of  its  altitude  by  the  combined  sum  of  the  two 
bases  and  four  times  the  mid-section. 


Given:  The  prismatoid  ACD-EFGKT,  with  bases  b  and  .B, 
mid-section  Jf,  altitude  H,  and  volume  F. 

To  Prove  :    V  =  %  H\b  +  B  +  ±M]. 

Proof :  Through   any   point,  P,  in   the   mid-section,  pass 
planes,  each  containing  an  edge  of  the  solid. 

These  planes  will  divide  the  prismatoid  into  pyramids : 
I.     The  pyramid  P-ACD,  whose  vertex  is  P,  base  6,  alti- 
tude \  H. 

Of  this  pyramid,  v1  =  J  b  •  H  (612). 

II.     The  pyramid  P-EFGKT,  whose  vertex  is  P,  base  .B, 
altitude  J  H. 

Of  this  pyramid,          vz  =  J  BH. 


POLYHEDRONS  353 

III.     Several  pyramids,  like  P-AEF,  whose  combined  vol- 
ume =  |  M  -  //,  as  we  shall  now  prove.     AE  =  2  AR  (?). 
.'.    AAEF=4AARS                                  (376). 
.*.  pyramid  P-AEF  =  4  -  pyramid  P-ARS           (614). 
But  pyramid  P-ARS  =  ^  A  PRS  •  |  J/=  J  A  PRS  -  II    (611). 

That  is,    pyramid  P-AEF  =  |  A  PRS  -  II. 
Hence,  for  the  sum  of  all  such  pyramids  in  the  prismatoid, 
VS  =  ±M-H  (Ax.  4). 

By  addition,  F  =  £  bH  +  £  7*#  +  £  3f //     (Ax.  2). 

Or,  V=\H[b  +  ^  +  43f].  Q.E.D. 


Ex.  1.  A  prismatoid  has  an  upper  base  5  sq.  in.,  a  lower  base  11  sq. 
in.,  a  mid-section  8  sq.  in.,  and  an  altitude  9  in.  Find  the  volume. 

Ex.  2.  How  does  the  mid-section  of  a  prism  compare  with  the  base  ? 
the  mid-section  of  a  pyramid?  of  a  cube-? 

Ex.  3.  Will  the  formula  for  the  volume  of  a  prismatoid  give  the  vol- 
ume of  a  cube?  Will  it  give  the  formula  for  the  volume  of  a  prism? 

Ex.  4.  Reduce  the  prismatoid  formula  to  a  formula  for  the  volume 
of  a  parallelepiped. 

Ex.  6.  Derive  the  formula  for  the  volume  of  a  pyramid  from  the 
prismatoid  formula. 

Ex.  6.  Is  a  prism  a  prismatoid ?  is  a  pyramid ?  is  a  truncated  prism? 
is  the  frustum  of  a  pyramid?  is  a  parallelepiped? 

Historical  Note.  The  prismatoid  formula  was  discovered  by  a  German, 
E.  F.  August,  in  the  latter  part  of  the  nineteenth  century.  Its  importance 
in  the  mensuration  of  polyhedrons  was  recognized  at  once  by  mathema- 
ticians throughout  the  world. 

REGULAR  AND  SIMILAR  POLYHEDRONS 

629.  A  regular  polyhedron  is  a  polyhedron  whose  faces  are 
equal  regular  polygons  and  whose  polyhedral  angles  are  all 
equal. 

Similar  polyhedrons  are  polyhedrons  which  have  the  same 
number  of  faces  similar  each  to  each  and  similarly  placed, 
and  which  have  their  homologous  polyhedral  angles  equal. 


354 


BOOK   VII.     SOLID   GEOMETRY 


PROPOSITION  XXIX.     THEOREM 

630.  There  can  exist  no  more  than  five  kinds  of  regular 
polyhedrons. 

Proof:   The  faces  must  be  equilateral  A,  squares,  regular 

pentagons,  or  some  other  regular  polygons.  (629.) 

There  must  be  at  least  three  faces  at  each  vertex.  (565.) 

The  sum  of  the  face  A  at  each  vertex  is  <  360°.  (549.) 

I.  Each  Z  of  an  equilateral  A  =  60°  (?).     Hence  we  may 
form  a  polyhedral  Z  by  placing  3,  4,  or  5  equilateral  A  at  a 
vertex,  but  not  6   (?).     That  is,  only  three  regular  poly- 
hedrons can  be  formed  having  equilateral  triangles  for  faces. 

II.  Each  Z  of  a  square  =  90°  (?).     Hence  we  may  form 
a  polyhedral  Z  by  placing  3  squares  at  a  vertex ;  but  not  4. 
That  is,  only  one  regular  polyhedron  can  be  formed  having 
squares  for  faces. 

III.  Each  Z  of  a  regular  pentagon  =  108°  (155).     Hence 
we  may  form  a  polyhedral  Z  by  placing  3,  but  not  4  regular 
pentagons  at  a  vertex.     That  is,  only  one  regular  polyhedron 
can  be  formed  having  regular  pentagons  for  faces. 

IV.  Each  Z  of  a  regular  hexagon  =  120°  (?). 
.•.  no  polyhedral  Z  can  be  formed  by  hexagons      (?). 

Consequently  there  can  be  no  more  than  five  kinds  of 
regular  polyhedrons,  —  three  kinds  bounded  by  triangles, 
one  kind  by  squares,  and  one  by  pentagons.  Q.E.D. 

631.  The  names  of  the  regular  polyhedrons. 


TOTAL 

NUMBKR   OF 

NAMES 

NUMBER 

FACRB  AT 

KINDS  OF  FACES 

OK  FACES 

EACH  VERTEX 

Regular  tetrahedron 

4 

3 

Equilateral  triangles 

Regular  hexahedron  (cube) 

6 

3 

Squares 

Regular  octahedron 

8 

4 

Equilateral  triangles 

Regular  dodecahedron 

12 

3 

Regular  pentagons 

Regular  icosahedron 

20 

5 

Equilateral  triangles 

POLYHEDRONS 


355 


KKGUL.AR  CUHK  RKGUL.AR  RKUULAR  KKGULAK 

TETRAHEDRON  OCTAHEDRON        DODECAHEDRON     ICOSAHEDRON 


DIRECTIONS  FOR  CONSTRUCTION.  —  Mark  on  cardboard  larger  figures 
similar  to  the  drawings.  Cut  the  dotted  lines  half  through  and  the  solid 
lines  entirely  through.  Fold  along  the  dotted  lines,  closing  the  solids  up 
and  forming  the  figures.  Paste  strips  of  paper  along  the  edges. 


Historical  Note.  Pythagoras  knew  about  the  existence  of  all  the  regu- 
lar polyhedrons  except  the  dodecahedron.  This  was  discovered  it)  470  B.C. 
by  Hippasus,  who  having  boasted  of  his  discovery  was  drowned  by  the 
other  Pythagoreans.  The  regular  polyhedrons  were  supposed  to  have 
certain  magical  properties  and  their  study  was  greatly  emphasized. 

ROBBINS'S    NEW   SOLID    GEOM    — 8 


356  BOOK   VII.     SOLID   GEOMETRY 

PROPOSITION  XXX.    THEOREM 

632.   In  two  similar  polyhedrons : 

I.   Homologous  edges  are  proportional. 

n.  Homologous  faces  are  to  each  other  as  the  squares  of 
any  two  homologous  edges. 

III.  Total  surfaces  are  to  each  other  as  the  squares  of  any 
two  homologous  edges. 


Proof :  I.  Homologous  faces  are  similar          (629). 

AB         EC         CD         DH         AE         BF 


A'B'     B'C'     C'D'     D'H'     A'E'     B'F* 
Face  DG        im*        face  AH        ~AE' 


Face  D'G1      .z/jj'2      face  A'a®     ~ 
Face  DG        face  AH        face  GE 


Face  D'(?'  "  face  A'H'      face  GrEf 
Total  surface  of  A  G       face  DG       DH        AE 


_    ^ 
~ 


Total  surface  of  A'Q'     face  D'  G1     vTa       A'W2 

Or,  T:T'  =  e2;e'2.  Q.E.D. 

633.  COROLLARY.  If  a  pyramid  is  cut  by  a  plane  parallel  to 
the  base,  the  pyramid  cut  away  is  similar  to  the  original 
pyramid.  _  (629.) 

Ex.  1.  Show  that  the  theorem  of  624  holds  true  in  the  case  of  a  cube. 
Ex.  2.  Show  that  the  theorem  of  626  holds  true  in  the  case  of  a  cube. 
Ex.  3.  Show  that  624  and  626  are  true  of  a  regular  octahedron. 


POLYHEDRONS  357 

PROPOSITION  XXXI.     THEOREM 

634.  Two  similar  tetrahedrons  are  to  each  other  as  the  cubes 
of  any  two  homologous  edges. 


O' 


Given  :  Similar  tetrahedrons  O-ABC  and  O'-A'B'C'  ;  whose 
volumes  =  V  and  V'. 
To  Prove  :   v  :  V1  =  2s3 :  J7^73  =  etc. 

Proof :  Trihedral  Z  A  =  trihedral  Z  ^  (629). 

.     F  AB.AC.AO  (623) 


(632,  I). 

(Ax.  6). 
etc.     Q.E.D. 


'  V' 

A'B' 

.A'C' 

^A'O' 

AB 

AC 

AO 

A'B' 

A'C' 

A'O' 

AB 

AC 

AO 

A'  B' 

A'C' 

A'O1 

p 

V 

AB 

AB 

AB 

'  V' 

A'B' 

A'B' 

A'B' 

That  is, 

F  :  V' 

=  ABB  : 

A'B'& 

=  AC3  :  A'C' 

Or, 

S:T 

r'  =  e* 

:  e>  3. 

Ex.  1.  If,  in  the  figure  of  631,  AO=  4  in.  and  A'0'  =  l  in.,  what  is 
the  ratio  of  the  total  surfaces  of  the  tetrahedrons  ?  of  their  volumes  ? 

Ex.  2.  Two  homologous  edges  of  two  similar  polyhedrons  are  2  in. 
and  3  in.  What  is  the  ratio  of  their  total  surfaces  ? 

Ex.  3.  Two  homologous  edges  of  two  similar  tetrahedrons  are  2  in. 
and  5  in.  The  total  surface  of  the  less  is  28  sq.  in.,  and  its  volume  is  40 
cu.  in.  Find  the  area  of  the  total  surface  and  the  volume  of  the  other. 

Ex.  4.  Show  that  the  theorems  of  624  and  626  are  true  in  the  cases  of 
regular  dodecahedrons  and  regular  icosaliedrous. 


358  BOOK   VII.     SOLID   GEOMETRY 

PROPOSITION  XXXII.     THEOREM 

635.  Two  similar  polyhedrons  can  be  decomposed  into  the 
same  number  of  tetrahedrons  similar  each  to  each  and  sim- 
ilarly placed. 


Given  :     (?). 
To  Prove  :     (?). 

Proof:  Suppose  diagonals  drawn  in  every  face  of  AT^  ex- 
cept the  faces  containing  vertex  A,  dividing  the  faces  into  A. 
(The  figure  shows  only  8V.) 

Suppose  lines  drawn  from  A  to  the  several  vertices  of 
these  A.  (The  figure  shows  only  AS,  AV.) 

Obviously,  this  process  divides  the  solid  (by  planes)  into 
tetrahedrons,  each  of  which  has  a  vertex  at  A. 

Then   construct  homologous  lines  in   solid  A'T*.     There 
will  evidently  be  as  many  lines  in  AfTf  as  in  AT  and  as  many 
tetrahedrons,  and  these  will  be  similarly  placed. 
Now,  in  the  tetrahedrons  A-SVR  and  Af-SrVrRf, 
A  AYR  is  similar  to  A  A'V'R'  ; 
A  ASR  is  similar  to  A  A'S'R'  ; 
'  A  SVR  is  similar  to  A  Sf  VrRf  (318). 

AV         VR          VS         RS         AS 


A1 

Also 


AV   =    VS 
A'V'  ~  V'S1 


POLYHEDRONS  359 

Hence  A  ASV  is  similar  to  A  A' 81  V1  (308). 

Also  the   trihedral  A  R  and   Rf   are   equal ;   8  and   Sf  are 

equal  ;    V  and  Vr  are  equal,  etc.  (546). 

.-.  the  two  tetrahedrons  are  similar  (Def.  629). 

Furthermore,  after  removing  these  tetrahedrons,  the  re- 
maining polyhedrons  are  similar.  (Def.  629.) 

By  the  same  process  other  pairs  of  tetrahedrons  may  be 
removed  and  proved  similar,  and  the  process  may  be  con- 
tinued until  the  polyhedrons  are  completely  decomposed  into 
tetrahedrons  similar  each  to  each  and  similarly  placed.  Q.E.D. 

PROPOSITION  XXXIII.     THEOREM 

636.  The  volumes  of  two  similar  polyhedrons  are  to  each 
other  as  the  cubes  of  any  two  homologous  edges. 

Given:  Similar  polyhedrons  AT  and  A'T'  ;  volumes  V  and 
V'\  AB  and  A'R1,  any  two  homologous  edges.  (See  figure 
of  635.) 

To  Prove  :    (?). 

Proof:  These  solids  may  be  decomposed,  etc.  (635). 

Denote  the  volumes  of  tetrahedrons  of  AT  by  w,  x,  y,  z, 
etc.;  of  A'T'  by  w',  xf,  y',  z' ,  etc. 

rpi                w       Alf      x       AR3       y       AI?  /i?oi\ 

Ihen         —  ==r^;  -7= »;  -£•  = ^;  etc.        (b34). 

v*     A'R'*    *'     A'R'3    y'     A'R'3 

(Ax.  1). 

(291) 

(Ax.  6).     Q.E.D. 
A'R' 

637.  COROLLARY.     The  volumes  of  two  similar  pyramids  are 
to  each  other  as  the  cubes  of  their  altitudes.     (Explain.) 


Hence 
Therefore      w  +  x 

w'     x'     y' 
-h?/4-etc.   _w 

w'  +  x> 
That  is. 

'  +y'  H-  etc.     w1 

3 

360 


BOOK   VII.     SOLID   GEOMETRY 


ORIGINAL  EXERCISES 

1.  What  plane  through  the  vertex  of  a  given  tetrahedron  will  divide 
it  into  two  equal  parts?     Prove. 

2.  The  area  of  the  base  of  any  pyramid  is  less  than  the  sum  of  the 
lateral  faces. 

[Draw  the  altitudes  of  the  lateral  faces  and  the  projections   of  the 
altitudes  upon  the  base.] 

3.  Three  of  the  edges  of  a  parallelepiped  that  meet 
in  a  point  are  also  the  lateral  edges  of  a  pyramid. 
What  part   of  the   parallelepiped   is  this  pyramid? 

4.  A  plane  is  passed  containing  one  vertex  of  a 
parallelepiped  and  a  diagonal  of  a  face  not  contain- 
ing that  vertex.     What  part  of   the    volume  of   the 
parallelepiped  is  the  pyramid  thus  cut  off  ? 

6.  Any  section  of  a  tetrahedron  made  by  a  plane 
parallel  to  two  opposite  edges,  is  a  parallelogram. 

Given :  Section  DEFG  II  to  OA  and  BC. 

To  Prove :  DEFG  is  a  O. 

Proof:  EF  is  II  to  OA,  DG  is  II  to  OA  (?).  Also 
DE  is  II  to  BC  and  GF  is  II  to  BC  (?),  etc. 

6.  The  three  lines  that  join  the  midpoints  of  the  opposite  edges  of  a 
tetrahedron  meet  in  a  point  and  bisect  one  another. 

Given :  LM,  PQ,  RS,  three  lines,  etc.  To  Prove  :  (?). 
Proof :  Join  PS,  SQ,  QR,  PR.    PS  is  II  to  and  = 
£  BC ;  RQ  is  II  to  and  =  \  BC.     (Explain.) 
.-.fig.  PSQRis&O  (?). 
Similarly,  discuss  LM  and  SR. 

7.  A  pyramid  having  one  of  the  faces  of  a  cube 
for  its  base  and  the  center  of  the  cube  for  its  vertex, 
contains  one  sixth  of  the  volume  of  the  cube. 

8.  A  plane  containing  an  edge  of  a  regular  tetra- 
hedron and  the  midpoint  of  the  opposite  edge : 

(a)  contains  the  medians  of  two  faces ; 

(b)  is  perpendicular  to  the  opposite  edge ; 

(c)  is  perpendicular  to  these  two  faces ;  M 

(d)  contains  two  altitudes  of  the  tetrahedron. 

9.  The  altitude  of  a  regular  tetrahedron  meets  the  base  at  the  point 
of  intersection  of  the  medians  of  the  base. 


ORIGINAL   EXERCISES 


361 


10.  The  altitude  of  a  regular  tetrahedron  =  £\/6  times  the  edge. 

11.  The  altitudes  of  a  regular  tetrahedron  meet  at  a  point. 

12.  The  lines  joining  the  vertices  of  any  tetrahedron  to  the  point  of 
intersection  of  the  medians  of  the  opposite  face  meet  in  a  point  that 
divides  each  line  into  segments  in  the  ratio  3 : 1. 

Given :  OM,  CR,  two  such  lines. 
To  Prove  :  The  four  such  lines  meet,  etc. 
Proof:  OM  and  CR  lie  in  the  plane  determined 
by  OC  and  point  D,  the  midpoint  of  AB. 
.:  OM  and  CR  intersect.     Draw  RM. 
DR  =\RO  1     (^       .    PR  _  RO 
DM=\  MC  \  '  DM~  MC 

.-.^M  is  II  to  OC  B 

.-.  A  DMR  and  DCO  are  similar  (?)  ;  and  DR:DO  =  RM:  OC.       (?) 
Thus  RM  =  £  OC.     (Explain.) 

Also  &  PRM  and  OPC  are  similar  (?). 

Hence  OP  :  PM  =  OC  :  RM  =  3:1.) 

And  CP:PR=OC:RM  =  3:l.\  (Explain.) 

Q.E.D. 

NOTE.     This  point  P  is  called  the  center  of  gravity  of  the  tetrahedron. 

13.  There  can    be   no   polyhedron    having    seven 
edges  and  only  seven. 

14.  The  planes  bisecting  the  dihedral   angles  of 
any  tetrahedron  meet  in  a  point  that  is  equally  distant 
from  the  faces. 

16.  The  lines  perpendicular  to  the  faces  of  any 
tetrahedron,  at  the  centers  of  the  circles  circumscribed 
about  the  faces,  meet  in  a  point  that  is  equally  distant  from  the  vertices. 

Proof :  RX  and  SY  are  loci  of  points,  etc.  (511). 

Plane  MN,  _L  to  AB  at  M,  the  midpoint  of  AB, 
is  the  locus  of  points,  etc. 

.-.  RX  and  SY  lie  in  MN  and  intersect  at  0,  etc. 

16.  If  a  plane  is  passed  through  the  midpoints 
of  the  three  edges  of  a  parallelepiped  that  meet  at 
a  vertex,  what  part  of  the  whole  solid  is  the  pyramid 
thus  cut  off? 

17.  The  plane  bisecting  a. dihedral  angle  of  a  tetrahedron  divides  the 
opposite  edge  into  two  segments  proportional  to  the  areas  of  the  faces 
that  form  the  dihedral  angle. 


362 


BOOK   VII.     SOLID   GEOMETRY 


18.  Two  tetrahedrons  are  similar  if  a  dihedral  angle  of  one  equals  a 
dihedral  angle  of  the  other  and  the  faces  forming  these  dihedral  angles 
are  respectively  similar. 

19.  If  from  any  point  within  a  regular  tetrahedron  perpendiculars  to 
the  four  faces  are  drawn,  their  sum  is  constant  and 

equal  to  the  altitude  of  the  tetrahedron.  |E 

• 

20.  Construct  a  regular  tetrahedron  upon  a  given 
edge. 

Construction:  Upon  AB,  construct  an  equilateral 
&ABC.  Erect  ED  _L  to  plane  of  &ABC,  at  Z>,  the 
center  of  circumscribed  O.  Take  V  on  ED  such  that 
AV  =  BV  =  CV  =  AB,etc. 

21.  Construct  a  regular  hexahedron  upon  a  given 


Construction:  Upon  AB  construct  a  square 
ABCD.  At  the  vertices  erect  _ls  =  to  AB  and  join  the 
extremities,  etc. 

22.  Construct  a  regular  octahedron  upon  a  given    . 
edge.  \ 

Construction:      Upon     AB     construct     a     square          D 
ABCD.      At  M,  the  center  of  the  square,  erect  XX' 
_L  to  plane  of  ABCD.     On  XX'  take  MV  =  MV  = 
MD.     Draw  the  edges  from  V  and  V. 

Statement  :    VV  is  a  regular  octahedron. 

Proof:  The  right  A  DM V,  DMC,  DMV  are 
equal.  (Explain.)  A 

Thus  the  12  edges  are  equal  and  the  8  faces  are 
equal.  (Explain.) 

Figures  AVCV,  DVBV,  ABCD  are  equal  squares 
(Explain.) 

Then,  pyramids  V-ABCD,  D-AVCV,  etc.,  are 
equal  and  the  6  polyhedral  angles  are  equal.  (Explain.)  .• 

23.  Pass  a  plane  through  a  cube  so  that  the  section  will  be 
hexagon. 

24.  Pass  planes  through  three  given  lines  in  space,  no  two  of  which 
are  parallel,  which  shall  inclose  a  parallelepiped. 

25     Find   the  lateral  area  and  the  total  area  of  a  regular  pyramid 
whose  slant  height  is  20  in.  and  whose  base  is  a  square,  1  ft.  on  a  side. 


ORIGINAL   EXERCISES 


363 


26.  Find  the  volume  of  a  pyramid  whose  altitude  is  18  in.  and  whose 
base  is  an  equilateral  triangle  each  side  of  which  is  8  in. 

27.  A  regular  hexagonal  pyramid  has  an  altitude1  of  9  ft.  and  each 
edge  of  the  base  is  6  ft.     Find  the  volume. 

28.  The  base  of  a  pyramid  is  an  isosceles  triangle  whose  sides  in  inches 
are  14,  25,  25,  and  the  altitude  of  the  pyramid  is  12  in.    Find  its  volume. 

29.  The  altitude  of  the  frustum  of   a  pyramid  is  25    in.,  and  the 
bases  are  squares  whose  sides  are  4  in.  and  10  in.,  respectively.     Find 
the  volume  of  the  frustum. 

30.  The  frustum  of  a  regular  pyramid  has  hexagons  for  bases  whose 
sides  are  5  in.  and  9  in.,  respectively.     The  slant  height  of  the  frustum 
is  14  in.     Find  its  lateral  area.     Find  its  total  area.  Q 

31.  The  altitude  of  a  regular  pyramid  is  15  in.,  and 
each  side  of  its  square  base  is  16  in.     Find  the  slant 
height,  the  lateral  edge,  the  total  area,  and  the  volume. 

OA*  =  O&  +  DA2  =  (15)2  +  (8)2  =  289. 
.-.A0  =  17.     OC2  =  OA 2  +  A  C*  =  289  +  64  =  353. 
.-.  OC  =  v/353  =  18.78+. 

32.  The  slant  height  of  a  regular  pyramid  is  39  ft.,  the 
altitude  is  36  ft.,  and  the  base  is  a  square.     Find  the 
lateral  area  and  the  volume. 

33.  The  lateral  edge  of  a  regular  pyramid  is  37  in. 
and  each  side  of  the  hexagonal  base  is  12  in.     Find  the 
slant  height,  the  lateral  area,  the  total  area,  and  the 
volume. 

In  rt.  A  A  CD,  CD  =  12,  A  C  =  6,     .-.  AD  =  6\/3. 
Inrt.  AvlCO,  CO  =  37,  AC  =  6,     .-.  .40  =  V1333. 
In  rt.  A  CDO,  CO  =  37,  CD  =  12,  .-.  OD  =  35,  etc. 

34.  Find  the   total  area  and  volume   of  a  regular 
tetrahedron  whose  edge  is  6  in. 

The  four  faces  are  equal  equilateral  &.     .-.  AO=AC 
=  3\/3  in.;  .-.  AD  =  V3  in.  and  CD  =  2V3  in. 

Hence  OD  =  2  V6  in.     Area  of  any  face  =  9V3  sq. 
in.,  etc. 

36.    Find   the  total    area  and  the  volume  of  a   regular  tetrahedron 
whose  edge  is  10  in. 

36.    Find  the  total  area  and  the  volume  of  a  regular  hexahedron  whose 
edge  is  8  in. 


864  BOOK   VII.     SOLID   GEOMETRY 

37.  Find  the  total  area  and  the  volume  of  a  regular 
octahedron  whose  edge  is  16  in. 

The  8  faces  are  equal  equilateral  A.  ^40  =  8V3.  In 
A  ADO,  one  finds  OD  =  8V2.  The  volume  of  the  octa- 
hedron =  the  volume  of  two  pyramids,  etc. 

38.  Find  the  total  area  and  the  volume  of  a  regular 
octahedron  whose  edge  is  18  in. 

39.  The  altitude  of  a  regular  pyramid  is  16  in.  and  each  side  of  the 
square  base  is  24  in.     Find  the  lateral  area  and  the  volume. 

40.  The  slant  height  of  a  regular  pyramid  is  16  in.  and  each  side  is 
an  equilateral  triangle  whose  side  is  20  \/3  in.     Find  the  total  area  and 
the  volume. 

41.  The  altitude  of  a  regular  pyramid  is  29  in.  and  its  base  is  a  regu- 
lar hexagon  whose  side  is  10  in.     Find  the  total  area  and  the  volume. 

42.  Find  the  total  area  and    the  volume  of   a   regular   tetrahedron 
whose  edge  is  18  in. 

43.  Find  the  total  area  and  the  volume   of   a    regular  octahedron 
whose  edge  is  20  in. 

44.  If  the  edge  of  a  regular  tetrahedron  is  e  in.,  show  that  the  total 
area  is  e2\/3  in.  and  the  volume  is  ^  e3V2  cu.  in. 

45.  If  the  edge  of  a  regular  octahedron  is  e  in.,  show  that  the  total  area 
is  2  e2v"3  sq.  in.  and  the  volume  is  \  e3V2  cu.  in. 

46.  A  pyramid  whose  base  is  a  square  9  in.  on  a  side,  contains  360  cu. 
in.     Find  its  height. 

47.  A  pyramid  has  for  its  base  a  hexagon  whose  side  is  7£  units  and 
the  pyramid  contains  675  cubic  units.     Find  the  altitude. 

48.  The  volume  of  a  regular  tetrahedron  is  144  V2   cu.   in.     Find   its 
edge. 

49.  The  volume  of  a  regular  octahedron  is  243  V2   cu.   in.     Find  its 


60.  The  volume  of  a  square  pyramid  is  676  cu.  in.  and  the  altitude  is 
1  ft.     Find  the  side  of  the  base.     Find  the  lateral  area. 

61.  The  altitude  of  the  Great  Pyramid  is  480  ft.  and  its  base  is  764 
ft.  square.     It  is  said  to  have  cost  $  10  a  cubic  yard  and  f  3  more  for  each 
square  yard  of  lateral  surface  (considered  as  planes).    What  was  the  cost? 

62.  The  total  surface  of  a  regular  tetrahedron  is  324\/3sq.  in.    Find 
its  volume. 

63.  The  base  of  a  pyramid  is  an  isosceles  right  triangle  whose  hypot- 
enuse is  8  in.     The  altitude  of  the  pyramid  is  15  in.     Find  the  volume. 


ORIGINAL   EXERCISES  365 

64.  Find  the  area  of  the  section  of  a  triangular  pyramid,  each  side  of 
whose  base  is  8  in.  and  whose  altitude  is  18  in.,  made  by  a  plane  parallel 
to  the  base  and  1  ft.  from  the  vertex. 

66.  The  altitude  of  a  frustum  of  a  pyramid  is  6  in.,  and  the  areas  of  the 
bases  are  20  sq.  in.  and  45  sq.  in.  Find  the  altitude  of  the  complete 
pyramid.  Find  the  volume  of  this  frustum  by  two  distinct  methods. 

66.  A  granite   inonument  in  the  form  of  a  frustum  of  a  pyramid, 
having  rectangular  bases  one  of  which  is  8  ft.  wide  and  12  ft.  long,  and 
the  other  6  ft.  wide,  is  30  ft.  high.      It  is  surmounted  by  a  granite 
pyramid  having  the  same  base  as  the  less  base  of   the  frustum,  and 
10  ft.  in  height.     Find  the  entire  volume  and  the  weight.     [1  cu.  ft.  of 
water  weighs  62|  Ib.  and  granite  is  3  times  as  heavy  as  water.] 

67.  If  a  square  pyramid  contains  40  cu.  in.  and  its  altitude  is  15  in., 
find  the  side  of  its  base. 

68.  A  church  spire  in  the  form  of  a  regular  hexagonal  pyramid  whose 
base  edge  is  8  ft.  and  whose  altitude  is  75  ft.  is  to  be  painted  at  the  rate 
of  18  ^  per  square  yard.     Find  the  cost. 

69.  Find  the  edge  of  a  cube  whose  volume  is  equal  to  the  volumes 
of  two  cubes  whose  edges  are  4  in.  and  6  in. 

60.  The  base  of  a  certain  pyramid  is  an  isosceles  trapezoid  whose 
parallel    sides  are  20  ft.  and    30    ft.   and   the  equal  sides  each  13  ft. 
Find  the  volume  of  the  pyramid  if  its  altitude  is  12  yd. 

61.  The  lateral  edge  of  the  frustum  of  a  regular  square  pyramid  is 
53  in.  and  the  sides  of  the  bases  are  10  in.  and  66  in.     Find  the  altitude, 
the  slant  height,  the  lateral  area,  and  the  volume. 

62.  The  sides  of  the  base  of  a  triangular  pyramid  in  inches  are  33, 
34,  65,  and  the  altitude  of  the  pyramid  is  80.     Find  its  volume. 

63.  The  sides  of  the  base  of  a  tetrahedron  in  inches  are  17,  25,  26,  and 
its  altitude  is  90.     Find  its  volume. 

64.  If  there  are  1£  cu.    ft.  in   a  bushel,  what  is  the  capacity   (in 
bushels)  of  a  hopper  in  the  shape  of  an  inverted  pyramid,  12  ft.  deep 
and  8  ft.  square  at  the  top  ? 

66.  In  the  corner  of  a  cellar  is  a  pyramidal  heap  of  coal.  The  base 
of  the  heap  is  an  isosceles  right  triangle  whose  hypotenuse  is  20  ft.  and 
the  altitude  of  the  heap  is  7  ft.  If  there  are  35  cu.  ft.  in  a  ton  of  coal, 
how  many  tons  are  there  in  this  heap? 

66.  How  many  cubic  yards  of  earth  must  be  removed  in  digging  an 
artificial  lake  15  ft.  deep,  whose  base  is  a  rectangle  180  ft.  x  20  ft.  and 
whose  top  is  a  rectangle  216  ft.  x  24  ft.  ?  [The  frustum'  of  a  pyramid.] 


366  BOOK  VII.     SOLID   GEOMETRY 

67.  One  pair  of  homologous  edges  of   two  similar  tetrahedrons  are 
3  ft.  and  5  ft.     Find  the  ratio  of  their  surfaces  ;  of  their  volumes. 

68.  A  pair  of  homologous  edges  of  two  similar  polyhedrons  are  5  in. 
and  7  in.     Find  the  ratio  of  their  surfaces  ;  of  their  volumes. 

69.  The  edge  of  a  cube  is  3  in.     What  is  the  edge  of  a  cube  twice  as 
large  ?    four  times  as  large  ?    half  as  large  ? 

70.  An  edge  of  a  tetrahedron  is  6  ft.     What  is  tjie  edge  of  a  similar 
tetrahedron  three  times  as  large?    eight  times  as  large?    nine  times  as 
large  ?    one  third  as  large  ? 

71.  An  edge  of  a  regular  icosahedron  is  3  in.     What  is  the  edge  of 
a  similar  solid  five  times  as  large?    ten  times  as  large?    fifty  times  as 
large?    a  thousand  times  as  large? 

72.  The  edges  of  a  trunk  are  2  ft.,  3  ft.,  5  ft.     Another  trunk  is  twice 
as  long  (the  other  edges  2  ft.  x  3  ft.).     How  do  their  volumes  compare? 
A  third  trunk  has  each  dimension  double  those  of  the  first.     How  does 
its  volume  compare  with  the  first  ?     How  do  their  surfaces  compare  ? 

73.  If  the  altitude  of  a  certain  regular  pyramid  is  doubled,  but  the 
base   remains  unchanged,  how  is  the  volume  affected?    If   each  edge 
of  the  base  is  doubled,  but  the  altitude  is  unchanged,  how  is  the  volume 
affected  ?    If  the  altitude  and  each  edge  of  the  base  are  all  doubled,  how 
is  the  volume  affected  ? 

74.  A  contractor  agrees  to  build  a  dam  60  ft.  long,  15  ft.  high,  11  ft. 
wide  at  the  bottom  and  7  ft.  wide  at  the  top  for  $  8.25  a  cubic  yard. 
Find  his  profit  if  it  costs  him  only  $2000. 

75.  A  pyramid  is  cut  by  a  plane  parallel  to  the  base  and  bisecting  the 
altitude.     What  part  of  the  entire  pyramid  is  the  less  pyramid  cut  away 
by  this  plane  ? 

76.  The  volume  of  a  certain  pyramid,  one  of  whose  edges  is  7  in.,  is 
686  cu.  in.     Find  the  volume  of  a  similar  pyramid  whose  homologous 
edge  is  8  in. 

77.  A  certain  polyhedron  whose  shortest  edge  is  2  in.  weighs  40  Ib. 
What  is  the  weight  of  a  similar  polyhedron  whose  shortest  edge  is  5  in.? 

78.  An  edge  of  a  polyhedron  is  5  in.  and  the  homologous  edge  of  a 
similar  polyhedron  is  7  in.    The  entire  surface  of  the  first  is  250  sq.  in.  and 
its  volume  is  375  cu.  in.    Find  the  entire  surface  and  volume  of  the  second. 

79.  A  berry  box,  sold  to  contain  a  quart  of  berries,  is  in  the  form  of 
the  frustum  of  a  pyramid  5  in.  square  at  the  top,  4|  in.  square  at  the 
bottom,  and  2£  in.  deep.     If  a  U.S.  dry  quart  contains  67.2  cu.  in.,  does 
this  box  contain  more  or  less  than  a  quart? 


BOOK   VIII 


CYLINDERS  AND   CONES 
CYLINDERS 

638.  A  cylindrical  surface  is  a  surface  generated  by  a 
moving  straight  line  which  continually  intersects  a  given 
curved  line  in  a  plane,  and  which  is  always  parallel  to  a 
given  straight  line  not  in  the  plane  of  the  curve. 

The  generating  line  is  the  generatrix.  The  directing 
curve  is  the  directrix. 

An  element  of  a  cylindrical  surface  is  the  generating  line 
in  any  position. 


CYLINDRICAL 
SURFACE 


RIGHT  CIRCULAR      OBLIQUE 
CYLINDER  CYLINDER 


CYLINDER 
OF  REVOLUTION 


639.    A  cylinder  is  a  solid  bounded  by  a  cylindrical  sur- 
face and  two  parallel  planes. 

The  bases  of  a  cylinder  are  the  parallel  plane  sections. 

The  lateral  area  of  a  cylinder  is  the  area  of  the  cylindrical 
surface  included  between  the  planes  of  the  bases. 

The  total  area  of  a  cylinder  is  the  sum  of  the  lateral  area 
and  the  areas  of  the  bases. 

The   altitude  of   a  cylinder  is  the  perpendicular  distance 
between  the  planes  of  the  bases. 

367 


368  BOOK   VIII.     SOLID   GEOMETRY 

640.  A  right  cylinder  is  a  cylinder  whose  elements  are  per- 
pendicular to  the  planes  of  the  bases. 

A  circular  cylinder  is  a  cylinder  whose  base  is  a  circle. 

An  oblique  cylinder  is  a  cylinder  whose  elements  are  not 
perpendicular  to  the  planes  of  the  bases. 

A  right  circular  cylinder  is  a  right  cylinder  whose  base  is 
a  circle. 

A  cylinder  of  revolution  is  a  cylinder  generated  by  the 
revolution  of  a  rectangle  about  one  of  its  sides  as  an  axis. 

Similar  cylinders  of  revolution  are  cylinders  generated  by 
similar  rectangles  revolving  on  homologous  sides. 

641.  A  right  section  of  a  cylinder  is  a  section  made  by  a 
plane  perpendicular  to  all  the  elements. 

A  plane  is  tangent  to  a  cylinder  if.  it  contains  one  element 
of  the  cylindrical  surface  and  only  one,  however  far  it  may 
be  extended. 

A  prism  is  inscribed  in  a  cylinder  if  its  lateral  edges  are 
elements  of  the  cylinder  and  the  bases  of  the  prism  are  in- 
scribed in  the  bases  of  the  cylinder. 

A  prism  is  circumscribed  about  a  cylinder  if  its  lateral 
faces  are  tangent  to  the  cylinder  and  the  bases  of  the  prism 
are  circumscribed  about  the  bases  of  the  cylinder. 

PRELIMINARY   THEOREMS 

642.  THEOREM.     Any  two  elements  of  a  cylinder  are  paral- 
lel and  equal.  (495  and  509.) 

643.  THEOREM.     A  line  drawn  through  any  point  in  a  cylin- 
drical surface,  parallel  to  an  element,  is  itself  an  element. 

(Ax.  13.) 

644.  THEOREM.     A  right  circular  cylinder  is  a  cylinder  of 
revolution. 

Ex.  If  a  plane  is  defined  as  a  surface  generated  by  a  moving  straight 
line,  what  is  the  directrix  ? 


CYLINDERS 


369 


THEOREMS  AND   DEMONSTRATIONS 

PROPOSITION  I.     THEOREM 

645.     Every  section  of  a  cylinder  made  by  a  plane  contain- 
ing an  element  is  a  parallelogram. 


" 


Given  :  Cylinder  AB ;  plane  CE  containing  element  CD. 
To  Prove  :   CE  is  a  O. 

Proof  :  At  E  draw  EF II  to  CD  in  plane  CE.  Also,  EF  is 
an  element  of  the  cylinder.  (643.) 

.-.  EF  is  the  intersection  of  the  plane  and  the  cylindrical 
surface.  (466.) 

Also  OF  is  II  to  DE  (-184). 

.-.  CDEFis&CJ  (120). 

PROPOSITION  II.     THEOREM 
646.     The  bases  of  a  cylinder  are  congruent. 


Given:  (?).     To  Prove  :  (?). 

Proof :  Suppose  2?,  5,  T  three  points  in  the  perim.  of  base  AC. 


(642). 


370  BOOK  VIII.     SOLID   GEOMETRY 

Draw  elements  RRr,  SSr<  Tf . 
Also  draw  RS,  ST,  RT,  R'S',  S'T',  R'T' 
RRf  =  and  is  II  to  88r 
RR1  =  and  is  II  to  Tl* 
88r  =  and  is  II  to  TTf 

.-.  RSf  is  a  O,  RT1  is  a  O,  ST1  is  a  O  (129). 

.'.RS  =  R'S';  8T=8rT1-,  RT=R'T'  (124). 

\'  /  * 

.•.  base  AC  may  be  placed  upon  base  BD  so  that  R,  S,  and 
T  coincide  with  Rf,  s',  and  r',  respectively.  But  S  is  any 
point  on  the  boundary ;  hence  every  point  on  the  boundary  of 
AC  will  coincide  with  a  corresponding  point  on  the  boundary 

of  BD. 

.-.  base  AC  ^  base  BD  (Def.  26). 

Q.E.D. 

647.  COROLLARY.     Parallel   plane   sections   of  a   cylinder 
(cutting  all  the  elements)  are  congruent. 

648.  COROLLARY.    Every  section  of  a  right  cylinder  made  by 
a  plane  containing  an  element  is  a  rectangle. 

649.  COROLLARY.    The  line  joining  the  centers  of  the  bases 
of  a  circular  cylinder  is  equal  and  parallel  to  an  element. 

650.  COROLLARY.    All  sections  of  a  circular  cylinder  parallel 
to  its  bases  are  equal  circles,  and  the  straight  line  joining  the 
centers  of  the  bases  passes  through  the  centers  of  all  the 
parallel  sections. 

To  Prove  :  HF  =  any  other  II  section  and 
C  is  its  center. 

Proof :  Pass  a  plane  AE  cutting  the  three 
planes  in  AG,  CF,  and  BE. 

DF  is  a  cylinder  (Def.  639). 


That  is,  HF  is  a  O  and  equal  to  any  other  section 


CYLINDERS 


371 


Now                              AG  is  II  to  CF  (484). 

And  A  K  is  II  to  GE  (649). 

.-.  ^FisaO  (120). 

And                                  AG  =  CF  (124). 

But  JIF  is  a  O  and  F  is  any  point  on  it.  Hence  C  is 

equally  distant  from  all  points  on  HF,  and  is,  therefore, 

the  center.  (179.) 

Q.E.D. 

PROPOSITION  III.     THEOREM 

651.  THEOREM.  If  a  regular  prism  is  inscribed  in,  or  circum- 
scribed about,  a  right  circular  cylinder  and  the  number  of  sides 
of  the  base  is  indefinitely  increased,  the  lateral  area  of  the 
cylinder  is  the  limit  of  the  lateral  area  of  the  prism. 


Given :  A  regular  prism  inscribed  in  and  a  regular  prism 
circumscribed  about  a  right  circular  cylinder  ;  the  lateral  area 
of  the  cylinder  =  L,  and  of  the  prisms,  Lt  and  Lc,  respectively. 

To  Prove :  That  as  the  number  of  sides  of  the  bases  of 
the  prisms  is  indefinitely  increased,  L  is  the  limit  of  both 
Li  and  Lc. 

Proof :  If  the  number  of  sides  of  the  bases  of  the  prisms  is 
indefinitely  increased,  their  perimeters  will  approach  the 
circumference  of  the  base  of  the  cylinder  as  a  limit.  (?.) 

Hence,  it  is  obvious  that  the  lateral  area  of  the  cylinder 
is  the  limit  of  the  lateral  area  of  either  prism.  Q.E.D. 

ROBBINS'S   NEW   SOLID   GEOM. 9 


372  BOOK  VIII.     SOLID   GEOMETRY 

PROPOSITION  IV.    THEOREM 

652.  If  a  prism  having  a  regular  polygon  for  a  base  is  in- 
scribed in,  or  circumscribed  about,  any  circular  cylinder  and 
the  number  of  the  sides  of  the  base  of  the  prism  is  indefinitely 
increased,  the  volume  of  the  cylinder  is  the  limit  of  the  volume 
of  the  prism. 


Given:  (?).    To  Prove:  (?). 

Proof :  If  the  number  of  sides  of  the  base  of  either  prism 
is  indefinitely  increased,  the  area  of  the  base  of  the  prism 
approaches  the  area  of  the  base  of  the  cylinder.  (424,  II.) 

.*.  it  is  obvious  that  the  volume  of  the  cylinder  is  the  limit 
of  the  volume  of  either  prism.  Q.E.D. 


Ex.  1.  If  the  cylindrical  surface  of  a  cylinder  is  cut  along  an  element, 
and  this  surface  is  placed  in  coincidence  with  a  plane,  what  plane 
geometrical  figure  will  it  become  9 

Ex.  2.     What  two  lines  determine  the  size  of  a  right  circular  cylinder? 

Ex.  3.  What  is  the  locus  of  all  points  2  in.  from  a  circular  cylindrical 
surface?  Is  the  answer  to  this  question  affected  by  the  radius  of  the 
given  surface?  If  so,  explain. 

Ex.  4.  From  a  log  36  in.  in  diameter  at  its  less  end,  is  to  be  cut  the 
largest  prismatic  piece  of  timber  possible,  having  square  ends.  Find  the 
side  of  this  square. 

Ex.  6.  A  lead  pencil  whose  ends  are  regular  hexagons  was  cut  from  a 
cylindrical  piece  of  wood,  with  the  least  waste  of  wood.  If  the  original 
piece  was  8  in.  long  and  \  in.  in  diameter,  find  the  volume  of  the  pencil. 


CYLINDERS 


373 


PROPOSITION  V.     THEOREM 

653.   The  lateral  area  of  a  right  circular  cylinder  is  equal  to 
the  product  of  the  circumference  of  the  base  by  an  element. 


Given  :  A  right  circular  cylinder,  the  circumference  of 
whose  base  =  O,  and  whose  element  =  E. 

To  Prove  :  Lateral  area  L  =  C-E. 

Proof  :  Inscribe  in  the  cylinder  a  regular  prism,  the 
perimeter  of  whose  base  is  P,  whose  lateral  edge  is  E,  and 
whose  lateral  area  is  Lr. 

Then  Lf  =  P-E  (?). 

If  the  number  of  sides  of  the  base  of  the  prism  is  indefi- 

nitely increased,  Lf  approaches  L  as  a  limit  (?). 

P  approaches  C  as  a  limit  (?). 

.-.  L  =  C-E  (?).     Q.E.D. 

654.    COROLLARY.     Area  of  a  right  circular  cylinder: 


(Where  L  =  lateral  area,  H  =  altitude,  E 
T  =  total  area.) 


radius  of  base,  and 


NOTE.  The  lateral  area  of  an  oblique  circular  cylinder 
equals  the  product  of  the  perimeter  of  a  right  section  of  the 
cylinder  by  an  element.  The  right  section  of  an  oblique  cir- 
cular cylinder  is  not  a  circle.  The  right  section  of  an  inscribed 
prism,  having  a  regular  polygon  for  a  base,  is  not  a  regular 
polygon.  Hence  the  proof  of  this  theorem  is  omitted. 


374  BOOK  VIII.     SOLID   GEOMETRY 

PROPOSITION  VI.     THEOREM 

655.    The  volume  of  a  circular  cylinder  is  equal  to  the 
product  of  its  base  by  its  altitude. 


Given :   A  circular  cylinder  whose  base  =  #,  altitude  =  IT, 
and  volume  =  F. 

To  Prove :   v  =  B  •  H. 

Proof :   Inscribe  a  prism  having  a  regular  polygon  for  its 
base,  whose  base  =  B1  and  volume  =  V1. 

Then  V1  =  Br  -  H  (?). 

If  the  number  of  sides  of  the  base  of  the  prism  is  indefi- 
nitely increased,  v'  approaches  F  as  a  limit  (?). 
B1  approaches  B  as  a  limit  (?). 
B1  -  H  approaches  B  •  H  as  a  limit. 

.-.  F  =  B  .H  (229).     Q.E.D. 

656.    COROLLARY.     Volume  of  a  circular  cylinder: 
r=  irR2jy. 

(Where  F  =  volume,  H—  altitude,  and  R  =  radius  of  base.) 


Ex.  1.  Find  the  lateral  area  and  the  total  area  of  a  cylinder  whose 
altitude  is  15  in.  and  radius  14  in. 

Ex.  2.  How  many  square  inches  of  tin  are  required  to  make  a  cylin- 
drical pipe  10 J  in.  in  diameter  and  8  ft.  long? 

Ex.  3.  What  is  the  capacity  of  a  cylindrical  pail  1  ft.  high  and  9  in. 
in  diameter? 


CYLINDERS 


375 


PROPOSITION  VII.    THEOREM 

657.    Of  two  similar  cylinders  of  revolution: 

I.  The  lateral  areas  are  to  each  other  as  the  squares  of 
their  altitudes  or  as  the  squares  of  the  radii  of  their  bases. 

II.  The  total  areas  are  to  each  other  as  the  squares  of 
their  altitudes  or  as  the  squares  of  the  radii  of  their  bases. 

III.  The  volumes  are  to  each  other  as  the  cubes  of  their 
altitudes  or  as  the  cubes  of  the  radii  of  their  bases. 


Given :  Two  similar  cylinders  of  revolution  whose  lateral 
areas  =  L  and  I ;  total  areas  =  T  and  t ;  volumes  =  V  and  v  ; 
altitudes  =  H  and  A,  and  radii  are  R  and  r. 
To  Prove  :  I.    L  :  1=  H2  :  A2  =  R2  :  r2. 

II.    T:  t=H2:  h2  =  R2  :  r*. 
III.    F  :  v=  Hs  :  A3  =  #3  :  r3. 

Proof :  The  generating  rectangles  are  similar. 

.-.  H:  h=R  :  r 

Hence  H+R_H_R 

h  -\-  r       h 

U       ff       TT      TT       rr2       7?2 

7= 


(Def.  640.) 
(291). 


2  Trrh 


RH 
rh 


TT 


r      n 
R     H+R 
r 


r 
H 

A  = 


A2      r2 


__  (  .         fi. 

'       ~~ 


7rR2H 


91 

irr2h 


R2Ff 

97 

rzh 


TTT  ___  _  __ 

-  ~  -  *         "  --       ~    *  -  -  ~~~ 


9 

r2 


H 

T" 

h 


H 


19  -  -      * 

h2      h      h6     r3 


Q.E.D. 


376  BOOK   VIII.     SOLID   GEOMETRY 

ORIGINAL  EXERCISES   (NUMERICAL) 

TT  =  3f.  1  bu.  =  2150.42  cu.  in.  1  gal.  =  231  cu.  in. 

In  a  cylinder  of  revolution  : 

1.  If  72  =  5  in.,  //  =  14  in.,  find  L ;   T-,    V. 

2.  If  R  =  7  m.,  If  =  10  in.,  find  L ;   T-    V. 

3.  If  72  =  4f  ft.,  77  =  18  ft.,  find  £ ;   T;    V. 

4.  If  72  =  6  in.,  L  =  792  sq.  in.,  find  //;   T-    V. 
6.   If  R  =  4  ft.,  T  =  352  sq.  ft.,  find  //;  Z;   V. 

6.  If  72  =  2  in.,  F  =  22  cu.  in.,  find  77;  L  ;   T7. 

7.  If  77  =  5.6  in.,  Z  =  352  sq.  in.,  find  R  ;   T7;    F. 

8.  If  77  =  9  in.,  T  =  440  sq.  in.,  find  R  ;  L  ;    F. 

9.  If  77  =  9£  in.,  F  =  66  cu.  in.,  find  R  ;  L  ;  71. 

10.  How  many  square  inches  of  tin  will  be  required  to  make  a  cylin- 
drical pail  10  in.    in    diameter   and   1    ft.  in   height,  without  any  lid? 
How  many  gallons  will  it  contain  ? 

11.  The  diameter  of  a  cylindrical  well  is  5|  ft.  and  the  water  is  14  ft. 
deep.    How  many  gallons  of  water  does  the  well  hold  ? 

12.  In  a  cylinder  of  revolution  generated  by  a  rectangle  30  in.  x  14 
in.  revolving  about  its  shorter  side  as  an  axis,  find  L ;   7*;    V. 

13.  In  a  cylinder  of  revolution  generated  by  the  rectangle  of  No.  12, 
revolving  about  its  longer  side  as  an  axis,  find  L  ;   T]    V. 

14.  A  cylindrical  vessel  9  in.  high,  closed  at  one  end,  required  361f 
sq.  in.  of  tin  in  its  construction.     Find  its  radius. 

15.  A  cylindrical  pail  12  in.  high  holds  exactly  2  gal.     Find  72. 

16.  How  many  cubic  feet  of  metal  are  there  in  a  hollow  cylindrical 
tube  42  ft.  long,  whose  outer  and  inner  diameters  are  10  in.  and  6  in.? 

17.  A  tunnel  whose  cross  section  is  a  semicircle  18  ft.  high  is  1  mi.  long. 
How  many  cubic  yards  of  material  were  removed  in  the  excavation  ? 

18.  An  irregular  stone  is  placed  in  a  cylindrical  vessel  a  in.  in  diame- 
ter and  partly  full  of  water.    The  water  rises  b  in.    Find  volume  of  stone. 

19.  A  rod  of  copper  18  ft.  long  and  2  in.  square  at  the  end  is  melted 
and  formed  into  a  wire  £  in.  in  diameter.     Find  the  length  of  the  wire. 

20.  Plow  many  miles  of  platinum  wira  ^z  in.  in  diameter  can  be  made 
from  a  cubic  foot  of  platinum? 

21.  If  a  cubic  foot  of  copper  weighs  550  lb.,  what  is  the  weight  of  a 
copper  wire  £  in.  in  diameter  and  5  mi.  long? 


CONES 


377 


CONES 

658.  A  conical  surface  is  a  surface  generated  by  a  moving 
straight  line  that  continually  intersects  a  given  curve  in  a 
plane,  and  passes  through  a  fixed  point  not  in  this  plane. 

The  generating  line  is  the  generatrix.  The  directing  curve 
is  the  directrix.  The  fixed  point  is  the  vertex  of  the  conical 
surface.  An  element  of  a  conical  surface  is  the  generating 
line  in  any  position. 


CONICAL 
SURFACE 


RIGHT 

CIRCULAR 

CONK 


FRUSTUM 
OF  A  CONE 


CONE  OF 
REVOLUTION 


OBLIQUE 
CONE 


659.  A  cone  is  a  solid  bounded  by  a  conical  surface  and  a 
plane  cutting  all  the  elements. 

The  base  of  a  cone  is  its  plane  surface. 

The  lateral  area  of  a  cone  is  the  area  of  the  conical  surface. 

The  total  area  of  a  cone  is  the  sum  of  the  lateral  area  and 
the  area  of  the  base. 

The  altitude  of  a  cone  is  the  perpendicular  distance  from 
the  vertex  to  the  plane  of  the  base. 

660.  A  circular  cone  is  a  cone  whose  base  is  a  circle. 

The  axis  of  a  circular  cone  is  the  line  drawn  from  the 
vertex  to  the  center  of  the  base. 

A  right  circular  cone  is  a  circular  cone  whose  axis  is  per- 
pendicular to  the  plane  of  the  base. 

An  oblique  circular  cone  is  one  whose  axis  is  oblique  to  the 
plane  of  the  base. 


378  BOOK   VIII.     SOLID   GEOMETRY 

A  cone  of  revolution  is  a  cone  generated  by  the  revolution 
of  a  right  triangle  about  one  of  the  legs  as  an  axis. 

Similar  cones  of  revolution  are  cones  generated  by  the 
revolution  of  similar  right  triangles  revolving  about  homol- 
ogous sides. 

The  slant  height  of  a  cone  of  revolution  is  any  one  of  its 
elements. 

661.  A  frustum  of  a  cone  is  the  portion  of  a  cone  between 
the  base  and  a  plane  parallel  to  the  base. 

The  altitude  of  a  frustum  of  a  cone  is  the  perpendicular 
distance  between  the  planes  of  its  bases.  The  slant  height 
of  a  frustum  of  a  cone  is  the  portion  of  an  element  included 
between  the  bases.  The  lateral  area  of  a  frustum  is  the  area 
of  its  curved  surface.  The  total  area  of  a  frustum  is  the 
sum  of  the  lateral  area  and  the  area  of  the  bases.  The  mid- 
section  of  a  frustum  is  the  section  made  by  a  plane  parallel 
to  the  bases  and  bisecting  the  altitude  and  the  slant  height. 

662.  A  plane  is  tangent  to  a  cone  if  it  contains  one  element 
of  the  conical  surface  and  only  one,  however  far  it  may  be 
extended. 

A  pyramid  is  inscribed  in  a  cone  if  its  base  is  inscribed  in 
the  base  of  the  cone,  and  its  vertex  is  the  vertex  of  the  cone. 

A  pyramid  is  circumscribed  about  a  cone  if  its  base  is  cir- 
cumscribed about  the  base  of  the  cone,  and  its  vertex  is  the 
vertex  of  the  cone. 

The  frustum  of  a  pyramid  is  inscribed  in,  or  circumscribed 
about,  the  frustum  of  a  cone  if  the  bases  of  -the  pyramid  are 
inscribed  in,  or  circumscribed  about,  the  bases  of  the  cone. 


Ex.  1.  Find  the  slant  height  of  a  right  circular  cone  whose  altitude 
is  8  in.  and  whose  radius  is  6  in. 

Ex.  2.     What  is  the  locus  of  all  points  3  in.  from  a  conical  surface  ? 

Ex.  3.  What  is  the  locus  of  all  lines  forming  a  given  angle  with  a 
given  line  at  a  given  point  in  the  line? 


CONES  379 

PRELIMINARY  THEOREMS 

663.  THEOREM.     The  elements  of  a  right  circular  cone  are 
all  equal.  (504,11.) 

664.  THEOREM.     A  right  circular  cone  is  a  cone  of  revo- 
lution. 

665.  THEOREM.     The  altitude  of  a  cone  of  revolution  is  the 
axis  of  the  cone. 

666.  THEOREM.     A  straight  line  drawn  from  the  vertex  of  a 
cone  to  any  point  in  the  perimeter  of  the  base  is  an  element. 

(39.) 

667.  THEOREM.     The  lateral  edges  of  a  pyramid  inscribed 
in  a  cone  are  elements  of  the  cone. 

668.  THEOREM.     The  lateral   faces  of  a  pyramid  circum- 
scribed about  a  cone  are  tangent  to  the  conical  surface. 

669.  THEOREM.     The  slant  height  of  a  regular  pyramid  cir- 
cumscribed about  a  right  circular  cone  is  the  same  as  the  slant 
height  of  the  cone. 

670.  THEOREM.     The  slant  height  of  the  frustum  of  a  regu- 
lar  pyramid  circumscribed   about   the   frustum   of    a   right 
circular  cone  is  the  same  as  the  slant  height  of  the  frustum 
of  the  cone. 

671.  THEOREM.     The  radius  of  the  mid-section  of  a  frustum 
of  a  right  circular  cone  is  equal  to  half  the  sum  of  the  radii 
of  the  bases. 

Proof :  The  radius  of  the  mid-section  is  the  median  of  a 
trapezoid  whose  bases  are  the  radii  of  the  bases  of  the 
frustum.  That  is,  m  =  J(B  +  r).  (139.) 

Q.E.D. 

Ex.  If  a  conical  surface  is  cut  along  an  element  and  the  surface  there 
placed  in  coincidence  with  a  plane,  what  geometrical  figure  does  the  sur- 
face become  ? 


380  BOOK   VIII.     SOLID   GEOMETRY 

672.  THEOREM.  If  a  regular  pyramid  is  inscribed  in,  or 
circumscribed  about,  a  right  circular  cone  and  the  number  of 
sides  of  the  base  is  indefinitely  increased,  the  lateral  area  of 
the  cone  is  the  limit  of  the  lateral  area  of  the  pyramid. 
(See  Fig.  A.) 

Demonstration  is  similar  to  that  of  651. 


FIG.  A  FIG.  B 

673.  THEOREM.     If  a  pyramid  having  a  regular  polygon  for 
a  base  is  inscribed  in,  or  circumscribed  about,  any  circular 
cone  and  the  number  of  sides  of  its  base  is  indefinitely  in- 
creased, the  volume  of  the  cone  is  the  limit  of  the  volume  of 
the  pyramid.     (See  Fig.  B.) 

Demonstration  is  similar  to  that  of  652. 

674.  THEOREM.     If  a  frustum  of  a  regular  pyramid  is  in- 
scribed in,  or  circumscribed  about,  the  frustum  of   a  right 
circular  cone  and  the  number  of  sides  of  the   bases  is  in- 
definitely increased,  the  lateral  area  of  the  frustum  of  the  cone 
is  the  limit  of  the  lateral  area  of  the  frustum  of  the  pyramid. 

675.  THEOREM.     If  the  frustum  of  a  pyramid  having  regular 
polygons  for  its  bases  is  inscribed  in,  or  circumscribed  about, 
a  frustum  of  any  circular  cone  and  the  number  of  sides  of 
the  bases  of  the  frustum  is  indefinitely  increased,  the  volume 
of  the  frustum  of  the  cone  is  the  limit  of  the  volume  of  the 
frustum  of  the  pyramid. 


CONES 


881 


THEOREMS  AND   DEMONSTRATIONS 

PROPOSITION  VIII.     THEOREM 

676.    Any  section  of  a  cone  made  by  a  plane  passing  through 
the  vertex  is  a  triangle. 


Given  :   Cone  O-AB ;  plane  OCD. 
To  Prove  :  Section  OCD  is  a  A. 

Proof :  Draw  straight  lines  OC,  OD,  in  plane  OCD. 

They  are  elements  (666). 

.*.  OC  and  OD  compose  the  intersection  of  the  plane  and 

the  conical  surface.  (466.) 

Also  CD  is  a  straight  line  (?). 

.-.  OCD  is  a  A  (23). 

Q.E.D. 

Ex.  1.  Can  the  plane  section  of  a  cone,  containing  the  vertex,  ever  be 
a  right  triangle?  Explain.  Can  it  be  an  isosceles  triangle?  Explain. 
What  kind  of  triangle  is  this  section,  in  general? 

Ex.  2.  What  is  the  locus  of  all  straight  lines  making  a  given  angle 
with  a  given  plane  at  a  given  point  in  the  plane  ? 

Ex.  3.  Does  every  cone  have  a  slant  height,  or  only  certain  kinds  of 
cones  ? 

Ex.  4.  If  a  circular  disk  is  held  between  a  source  of  light  and  a  wall, 
and  parallel  to  the  wall,  what  is  the  shape  of  the  shadow  on  the  wall? 
What  is  the  shape  of  the  shadow  region  between  the  light  and  the  wall? 
(Consider  the  source  of  light  a  point.) 


382 


BOOK  VIII.     SOLID  GEOMETRY 


PROPOSITION  IX.     THEOREM 

677.  Any  section  of  a  circular  cone  made  by  a  plane 
parallel  to  the  base  is  a  circle,  whose  center  is  the  intersec- 
tion of  the  plane  with  the  axis. 


Given :    Cone  O-AB ;    circle  C  its  base ;   section  Af  B1  II  to 
base,  and  axis  OC. 

To  Prove  :  A'B'  also  a  O,  whose  center  is  Cr. 
Proof :  Pass  planes  OCD,  OCE  intersecting  the  base  in  CD, 
CE  respectively,  and  the  section  in  C'D',  C'E'. 
In  A  OCD  and  OCE,  DrCr  is  II  to  DC ; 

C'E'  is  II  to  CE  (484). 

.-.  A  OC'D'  is  similar  to  A  OCD; 
A  OC'E'  is  similar  to  A  OCE 


OC 


But 

.•.  by  multiplying, 


CD 
C'D' 


OC 

cfEf 


CE 


CD         CE 
CD  =  CE 

C'D' 


(305). 
(313). 

(Ax.  1). 

(187). 
(Ax.  3). 


That  is,  all  points  on  the  boundary  of  A'B'  are  equally 
distant  from  C1. 

.-.  A'B'  is  a  O  whose  center  is  C          (179).     Q.E.D. 


Ex.     Any  section  of  a  circular  cone  parallel  to  the  base  is  to  the  base 
as  the  square  of  its  distance  from  the  vertex  is  to  the  altitude  of  the  cone. 

Proof :    A'B' :  AB  =  C^D'2 :  CD*  =  0~C'2 :  OC2.     (Explain.) 


CONES  383 

PROPOSITION  X.     THEOREM 

678.  The  lateral  area  of  a  right  circular  cone  is  equal  to 
half  the  product  of  the  circumference  of  the  base  by  the 
slant  height. 


Given:   Right  circular  cone   O-J.D,  the  circumference  of 
whose  base  =  C  and  whose  slant  height  =  s. 
To  Prove  :  Lateral  area  =  ^  C  •  s. 

Proof :  Circumscribe   a   regular  pyramid  and   denote  the 

lateral  area  by  Lr   and   the   perimeter   of   the   base   by   P. 

Slant  height  OA  =  s.  (663.) 

Then  Lf  =  ±  P  .  s  (598). 

Now  indefinitely  increase  the  number  of  sides  of  the  base 

of  the  pyramid  and  Lr  approaches  L  as  a  limit  (672), 

P  approaches  C  as  a  limit  (424,  I), 

£  P  •  s  approaches  ^  C  •  8  as  a  limit  (?). 

Hence  L  =  -J-  C  •  s  (229).     Q.E.D. 

679.     COROLLARY.     Area  of  a  right  circular  cone  : 

L  =  i  (2  TrR)s  =  trRs.          T  =  TrRs  +  irJB2  =  irR(s  +  .R). 

(Where  L  =  lateral  area,  T=  total  area,  *  =  slant  height,  and 
R  =  radius  of  base.)  

Ex.  1.  Of  a  right  circular  cone  whose  slant  height  is  15  in.  and  radius 
is  9  in.,  find  the  lateral  area,  the  total  area,  and  the  altitude. 

Ex.  2.  If  the  radius  of  a  right  circular  cone  is  8  in.  and  the  altitude 
is  15  in.,  find  the  slant  height,  the  lateral  area,  and  the  total  area. 


384  BOOK   VIII.     SOLID   GEOMETRY 

PROPOSITION  XI.     THEOREM 

680.  The  lateral  area  of  the  frustum  of  a  right  circular 
cone  is  equal  to  half  the  sum  of  the  circumferences  of  the 
bases  multiplied  by  the  slant  height. 


Given  :  Frustum  of  right  circular  cone,  whose  lateral  area 
is  L  ;  whose  slant  height  is  s  ;  and  the  circumferences  of 
whose  bases  are  C  and  c. 

To  Prove  :  L  =  J(c  +<?)•*• 

Proof  :  Circumscribe  a  frustum  of  a  regular  pyramid  and 
denote  its  lateral  area  by  L',  the  perimeters  of  its  bases  by  P 
and  p.  The  slant  height  of  frustum  of  pyramid  =  s  (670). 

Now  Z/  =  iJ-(P  +;?)•«  (605). 

Indefinitely  increase  the  number  of  the  sides  of  the  bases 
of  the  frustum  of  the  pyramid  and  L1  approaches  L  as  a  limit, 
P  approaches  C,  and  p  approaches  c  (?). 

J(P  +p)  •  8  approaches  ^(C+  c)  •  s  as  a  limit. 

Hence  L  =  %(C+e)  -  s  (229).     Q.E.D. 

681.  COROLLARY.  Area  of  the  frustum  of  a  right  circular 
C0ne  :  L  =  1(2  TTR  +  2  7rr>  =  ir(JB  +  r)s. 


L  =  7r(2  rri)s  =  2  irms. 
T=  TT(R  4-  r)s  +  7r/i2  +  -rrr2. 
.-.  T=Tr[(,R  +  r>  +  jR2  +  r2]. 

(Where  £,  T,  K,  r  and  «  denote  magnitudes  as  before,  and  m 
=  the  radius  of  the  midsection.) 


CONES 


385 


PROPOSITION  XII.     THEOREM 

682.     The  volume  of  a  circular  cone  is  equal  to  one  third 
the  product  of  the  area  of  the  base  by  the  altitude. 


Given:  Circular  cone  O-AV,  whose  volume  =  F;  area  of 
whose  base  =  B  ;  altitude  =  OE  =  H. 

To  Prove  :   F  =  J  B  -  H . 

Proof :  Circumscribe  (or  inscribe)  a  pyramid  having  a 
regular  polygon  for  its  base.  Denote  the  volume  of  the 
pyramid  by  F',  its  base  by  Br.  Its  altitude  =  OE  =  H  (491). 

Indefinitely  increase  the  number  of  the  sides,  etc. 
Then  V'  approaches  F  as  a  limit  (?). 

B'  approaches  B  as  a  limit  (424,  II). 

\  Bf  •  H  approaches  ^  B  •  H  as  a  limit. 
Hence  V=^B-H  (229).     Q.E.D. 

683.     COROLLARY.    Volume  of  a  circular  cone:  F=*TrR2JJ. 

(Where  F=  volume,  H  =  altitude,  and  It  =  radius  of  base.) 


Ex.  1.  Find  the  volume  of  a  circular  cone  whose  altitude  is  14  in.  and 
the  radius  of  the  base,  6  in. 

Ex.  2.  A  right  triangle,  whose  legs  are  15  in.  and  20  in.,  is  revolved 
about  the  lesser  leg  as  an  axis,  forming  a  cone  of  revolution.  Find  the 
lateral  area,  the  total  area,  and  the  volume.  Find  the  lateral  area,  the 
total  area,  and  the  volume  of  the  cone  formed  by  revolving  this  triangle 
about  the  greater  leg  as  an  axis. 


380  BOOK   VIII.     SOLID   GEOMETRY 

PROPOSITION   XIII.     THEOREM 

684.  The  volume  of  the  frustum  of  a  circular  cone  is  equal 
to  one  third  the  product  of  the  altitude  by  the  sum  of  the 
lower  base,  the  upper  base,  and  a  mean  proportional  between 
the  bases  of  the  frustum. 


Given :  A  frustum  of  any  circular  cone,  whose  volume  =  F, 
whose  bases  are  B  and  £>,  whose  altitude  is  H. 

To  Prove:   v=  ^  H\_B  +  b  +  V#  •  £>]. 

Proof :  Inscribe  (or  circumscribe)  a  frustum  of  a  pyramid 
having  regular  polygons  for  bases. 

Denote  its  volume  by  F',  bases  by  Bf  and  £>',  and  altitude 
by  H. 

Then  V  =  ^H  [Bf+bf  +  V#'  .  b']  (618). 

Indefinitely  increase,  etc.  Vf  approaches  F  as  a  limit  (?), 
Bf  and  b'  approach  B  and  b  respectively  as  limits  (?), 
V.B'  •  bf  approaches  V^  •  b  as  a  limit. 

%H[Bf  +  b'  +  ^/Bf  -  £>']  approaches  ^H[s+b  +  V#  •  &]. 

Hence  v=^H[B  +  l  +  Vz?  •  6]  (229).     Q.E.D. 

685.   COROLLARY.     Volume  of  the  frustum  of  a  circular  cone, 

-f  ^^2  _|_  V7TJ22   •  TTr2] 


(Where  F=  volume,  etc.) 


CONES 


387 


PROPOSITION  XIV.     THEOREM 

686.    Of  two  similar  cones  of  revolution : 

I.  The  lateral  areas  are  to  each  other  as  the  squares  of 
thek  altitudes,  or  as  the  squares  of   their  radii,  or  as  the 
squares  of  their  slant  heights. 

II.  The  total  areas  are  to  each  other  as  the  squares  of 
their  altitudes,  or  as  the  squares  of  their  radii,  or  as  the 
squares  of  their  slant  heights. 

III.  The  volumes  are  to  each  other  as  the  cubes  of  their 
altitudes,  or  as  the  cubes  of  their  radii,  or  as  the  cubes  of 
their  slant  heights. 

Given:  Two  similar  cones 
of  revolution,  whose  respec- 
tive lateral  areas  are  L  and  Z, 
total  areas  are  T  and  £,  vol- 
umes are  V  and  t;,  altitudes 
are  H  and  h,  radii  are  R  and 
r,  slant  heights  are  -S  and  s. 

'T          rr2         t.2 

To  Prove : 


I 


O  O 

r2      s2 


ii.  *.£.£.* 


O  O 

r2     s2 


III.    -1  =  ^=^  =  ^-. 


TJ*  -ry  O 

Proof:  Generating  A  are  similar  and  — =-=- 

h      r      s 


Hence 


L   ~ 

I        7TT8 


R+S_R_S_H 
r  -h  s      r      s      h 

7TRS=R      S^H      H=H2  =  B?. 
r     s      h      h      h2      r2 


(291). 
(Ax.  6). 


III.     v-  = 


T_7TR(R+S)==R 
R2 


H 


H      #3      I?3 


(?).       Q.E.D. 


ROBBINS'S    NEW    SOLID   GEOM. 10 


388  BOOK   VIII.     SOLID   GEOMETRY 

ORIGINAL  EXERCISES 
In  a  cone  of  revolution  : 

1.  If  //=  12  in.,  *  =  13in.,  find/?. 

2.  If  H  =  15  ft.,  R  =  8  ft.,  find  S. 

3.  If  R  =  18  cm.,  s  =  30  cm.,  find  H. 

4.  If  //  =  6  in.,                  s  =  10  in.,  find  R ;  L  ;   T;    V. 
6.   If  //  =  20  ft,              R  =  21  ft.,  find  s ;  L ;   71;   F. 

6.  If  #  =  7  m.,  *  =  25  m.,         find  //;  Z;   T;   F. 

7.  If  Z  =  4070  sq.,in.,     .s  =  37  in.,         find  R  ;  //;  T7;    F. 

8.  If  Z  =  46.64  sq.  in.,  72  =  2.8  in.,         find  s;  H;   T;   V. 

9.  If  Z  =  400  sq.  ft,       T  =  500  .q.  ft.,  find  s ;  H;  H  ;   F. 

10.  If  T  =  80  TT  sq.  in.,  R  =  5  in.,  finds;  #;  Z;    F. 

11.  If  T  =  10 TT  sq.  ft,  s  =  3  ft,  find  R;  //;  Z;    F. 

12.  If  F  =  462  cu.  in.,  R  =  21  in.,  find  //;  s ;  Z  ;   T. 

13.  If  F=8Tfycu.  ft,  7/=3ft.,  find  72;  *;  Z;   T. 

14.  What  would  be  the  cost  at  10 ^  a  square  foot  of  painting  a  conical 
church  steeple,  112  ft.  high  and  30  ft.  in  diameter  at  the  base? 

16.  The  sides  of  an  equilateral  triangle  are  each  12  in.  Find  the 
lateral  surface,  the  total  surface,  and  the  volume  of  the  solid  generated 
by  revolving  this  triangle  about  an  altitude  as  an  axis. 

16.  An  isosceles  right  triangle  whose  legs  are  each  8  in.  is  revolved 
about  the  hypotenuse  as  an  axis.     Find  the  total  surface  and  the  volume 
of  the  solid  generated. 

17.  The  sides  of  an  equilateral  triangle  are  each  10  in.     Find  the  total 
surface  and  the  volume  of  the  solid  generated  by  revolving  this  triangle 
about  one  of  its  sides  as  an  axis. 

18.  Find  the  volume  of  a  cone  of  revolution  whose  slant  height  is 
16  in.  and  whose  lateral  area  is  192  TT  sq.  in. 

19.  Find  the  lateral  area  of  a  cone  of  revolution  whose  altitude  is 
20  in.  and  whose  volume  is  240 TT  cu.  in. 

20.  How  many  bushels  are  there  in  a  conical  heap  of  grain  whose 
base  is  a  circle  35  ft.  in  diameter,  and  whose  height  is  25  ft.  ? 

21.  A  regular  hexagon  whose  side  is  6  in.  revolves  about  one  of  the 
longer  diagonals.     Find  the  surface  and  the  volume  of  solid  generated. 

22.  Find  the  volumes  of  the  right  circular  cones  inscribed  in  and 
circumscribed  about  a  regular  tetrahedron  whose  edge  is  a  m. 


ORIGINAL   EXERCISES  889 

23.  A  right  triangle  whose  legs  are  15  in.  and  20  in.  is  revolved  about 
the  hypotenuse  as  an  axis.     Find  the  surface  and  the  volume  of  the 
solid  generated. 

In  the  frustum  of  a  right  circular  cone : 

24.  If  //  =  8  in.,      R  =  10  in.,    r  =  4  in.,  find  s ;  L ;   T7;  V. 

25.  If  H  =  30  cm.,    s  =  34  cm.,  r  =  5  cm.,  find  R ;  L ;   T;    V. 

26.  If    s  =  19£  ft.,  72  =  10i  ft.,  r  =  3  ft.,  find  #;  Z;   T;   F. 

27.  How  many  square  feet  of  tin  are  required  to  make  a  funnel  2  ft. 
long,  if  the  diameters  of  the  ends  are  20  in.  and  56  in.,  respectively? 

28.  A  chimney  150  ft.  high  has  a  cylindrical  flue  3  ft.  in  diameter. 
The  bases  of  the  chimney  are  circles  whose  diameters  are  28  ft.  and  7  ft. 
Find  the  number  of  cubic  yards  of  masonry  in  the  chimney. 

29.  A  plane  is  passed  parallel  to  the  base  of  a  right  circular  cone  and 
|  the  distance  from  the  vertex  to  the  base.     Find  the  ratio  of  the  smaller 
cone  thus  formed  to  the  original  cone.     Compare  the  volume  of  the  less 
cone  with  the  frustum  formed. 

Original  cone  =  5J»    ?,      ^ 
Less  cone  23 

Hence          Original  cone  -  Less  cone  =  125  -  8  (?     ^ 
Less  cone  8 

30.  The  altitude  of  a  cone  of  revolution  is  12  in.     What  is  the  alti- 
tude of  the  frustum  of  this  cone  that  shall  contain  one  fourth  the  volume 
of  the  whole  cone  ? 

31.  The  altitudes  of  two  similar  cylinders  of  revolution  are  3  in.  and 

5  in.     What  is  the  ratio  of  their  lateral  areas?  of  their  total  areas?  of 
their  volumes? 

32.  The  altitudes  of  two  similar  cylinders  of  revolution  are  5  in.  and 

6  in.,  and  the  lateral  area  of  the  first  is  200  sq.  in.     Find  the  lateral  area 
of  the  second.     If  the  volume  of  the  first  is  500  cu.  in.,  what  is  the  volume 
of  the  second? 

33.  The  total  areas  of  two  similar  cones  of  revolution  are  24  TT  sq.  in. 
and  216  TT  sq.  in.  and  the  radius  of  the  first  is  3  in.     Find  the  radius  of 
the  second.     The  slant  height  of  the  first  is  5  in.     Find  the  lateral  area 
of  the  second.     Find  the  altitude  of  the  first  and  the  volume  of  the 
second. 

34.  The  volumes  of  two  similar  cones  of  revolution  are  27  IT  cu.  in.  and 
343  TT  cu.  in.     The  altitude  of  the  first  is  9  in.     Find  the  altitude  of  the 
second.     Find  the  radius  of  the  base  of  each. 


390  BOOK  VIII.     SOLID   GEOMETRY 

35.  A  cone  of  revolution  whose  radius  is  10  in.  and  altitude  20  in.,  has 
the  same  volume  as  a  cylinder  of  revolution  whose  radius  is  15  in.     Find 
the  altitude  of  the  cylinder. 

36.  A  cylinder  of  revolution  whose  radius  is  8  in.  and  altitude  30  in., 
is  formed  into  a  cone  of  revolution  whose  altitude  is  40  in.     Find  the 
radius  of  its  base. 

37.  The  heights  of  two  equivalent  cylinders  of  revolution  are  in  the 
ratio  of  4  :  9.     If  the  diameter  of  the  first  is  92  ft.,  what  is  the  diameter 
of  the  second? 

38.  A  cylinder  of  revolution  8  ft.  in  diameter  is  equivalent  to  a  cone 
of  revolution  7  ft.  in  diameter.     If  the  height  of  the  cone  is  16  ft.,  what 
is  the  height  of  the  cylinder? 

39.  Two  circular  cylinders  having  equa-1  altitudes  are  to  each  other  as 
their  bases. 

40.  Two  circular  cylinders  having  equal  bases  are  to  each  other  as 
their  altitudes. 

41.  Two  circular  cylinders  having  equal  bases  and  equal  altitudes  are 
equal. 

42.  Two  circular  cones  having  equal  altitudes  are  to  each  other  as 
their  bases. 

43.  Two  circular  cones  having  equal  bases  are  to  each  other  as  their 
altitudes. 

44.  Two  circular  cones  having  equal  bases   and   equal  altitudes  are 
equal. 

45.  If  the  altitude  of  a  right  circular  cylinder  is  equal  to  the  radius 
of  the  base,  the  lateral  area  is  half  the  total  area. 

46.  If  the  altitude  of  a  right  circular  cylinder  is  half  the  radius  of  the 
base,  the  lateral  area  is  equal  to  the  area  of  the  base. 

47.  If  the  slant  height  of  a  right  circular  cone  is  equal  to  the  diameter 
of  the  base,  the  lateral  area  is  double  the  area  of  the  base. 

48.  The  lateral  area  of  a  cone  of  revolution  is  equal  to  the  area  of  a 
circle  whose  radius  is  a  mean  proportional  between  the  slant  height  and 
the  radius  of  the  base. 

49.  The  lateral  area  of  a  cylinder  of  revolution  is  equal  to  the  area  of 
a  circle  whose  radius  is  a  mean  proportional  between  the  altitude  of  the 
cylinder  and  the  diameter  of  its  base. 

60.    What  relation  does  the  section  of  a  circular  cone  made  by  a  plane 
parallel  to  the  base  have  to  the  base  ?     Prove. 


ORIGINAL   EXERCISES  391 

61.  At  what  distance  from  the  vertex  of  a  right  circular  cone  whose 
altitude  is  h  must  a  plane  parallel  to  the  base  be  passed,  so  as  to  bisect 
the  lateral  area?  At  what  distance  must  it  be  passed  so  as  to  bisect  the 
volume? 

52.  What  does  the  volume  Tof  a  right  circular  cone  become,  if  the 
altitude  is  doubled  and  the  base  undisturbed?  Prove.  What  does  the 
volume  V  become  if  the  radius  of  the  base  is  doubled  but  the  altitude 
undisturbed  ?  Prove.  What  does  the  volume  become  if  both  radius  of 
base  and  altitude  are  doubled  ?  Prove. 

63.  The  intersection   of  two  planes  tangent   to  a  cylinder  is  a  line 
parallel  to  an  element. 

64.  The  intersection  of  two  planes  tangent  to  a  cone  is  a  line  through 
the  vertex. 

56.  One  straight  line  can  be  drawn  upon  a  cylindrical  surface  through 
a  given  point,  and  only  one. 

66.  If  two  cylinders  of  revolution  have  equivalent  lateral  areas,  their 
volumes  are  to  each  other  as  their  radii. 

57.  If  a  rectangle  is  revolved  about  its  unequal  sides  as  axes,  the  vol- 
umes of  the  two  solids  generated  are  inversely  proportional  to  the  axes, 
and  directly  proportional  to  the  radii  of  the  bases. 

68.  Show  that  the  formula  for  the  volume  of  a  circular  cone  can  be 
derived  from  the  formula  for  the  volume  of  a  frustum  of  a  circular  cone 
if  one  base  of  the  frustum  becomes  a  point. 

59.  Reduce  the  formula  for  the  volume  of  a  frustum  of  a  circular  cone 
if  the  radius  of  one  base  is  double  the  radius  of  the  other. 

60.  Could  you  prove  the  theorem  of  680  by  inscribing  a  frustum  of  a 
pyramid?     Could  you  prove  the  theorem   of  684  by  circumscribing  a 
frustum  of  a  pyramid?     Give  reasons  for  your  answer. 

61.  Could  you  prove  the  theorem  of  678  by  inscribing  a  pyramid? 
Could  you  prove  the  theorem  of   682  by  inscribing  a  pyramid?     Give 
reasons. 

62.  Pass  a  plane  tangent  to  a  circular  cylinder  and  containing  a  given 
element. 

Construction :  Draw  a  line  in  plane  of  base  tangent  to  the  base  at  the 
end  of  the  given  element,  etc. 

63.  Pass  a  plane  tangent  to  a  circular  cone  and  containing  a  given  ele- 
ment. 

64 .  Divide  the  lateral  surface  of  a  cone  of  revolution  into  two  equiva- 
lent parts  by  a  plane  parallel  to  the  base. 


392  BOOK   VIII.     SOLID   GEOMETRY 

65.  Pass  a  plane  tangent  to  a  circular  cylinder  and  through  a  given 
point  without  it. 

Construction:  From  the  point  draw  a  line  II  to  an  element,  meeting 
the  plane  of  the  base.  From  this  point  of  intersection  draw  a  line  tan- 
gent to  the  base  of  the  cylinder.  Through  the  point  of  contact  draw  an 
element,  etc. 

66.  Pass  a  plane  tangent  to  a  circular  cone  through  a  given  point 
without  it. 

Construction :  Connect  this  point  with  the  vertex  of  the  cone  and  pro- 
long this  line  to  meet  the  plane  of  the  base,  etc. 

67.  Find  the  locus  of  points  at  a  given  distance  from  a  given  straight  line. 

68.  Find  the  locus  of  points  equally  distant  from  two  given  points 
and  at  a  given  distance  from  a  straight  line.     Discuss. 

69.  Find  the  locus  of  points  at  a  given  distance  from  a  given  plane 
and  at  a  given  distance  from  a  given  line.     Discuss. 

70.  Find  the  locus  of  points  at  a  given  distance  from  a  given  cylin- 
drical surface  whose  generatrix  is  a  circle,  and  whose  elements  are  per- 
pendicular to  the  plane  of  the  circle. 

71.  Find  the  locus  of  a  point  at  a  given  distance  from  a  given  line 
and  equally  distant  from  two  given  planes.     Discuss. 

72.  Find  a  point  A',  at  a  given  distance  from  a  given  line  and  equally 
distant  from  three  given  points.     Discuss. 

73.  A  grain  elevator  in  the  form  of  a  frustum  of  a  right  circular  cone 
is  30  ft.  high,  and  the  radii  of  its  bases  are  12  ft.  and  8  ft.  respectively. 
If  a  bushel  contains  approximately  1  ^  cu.  ft.,  how  many  bushels  of  wheat 
will  this  elevator  hold? 

74.  A  certain  coffee  pot  is  7  in.  deep,  3£  in.  in  diameter  at  the  top,  and 
5  in.  at  the  bottom.     If  there  are  5  cups  in  a  quart,  how  many  cups  of 
coffee  will  this  coffee  pot  hold?     (Disregard  fractional  parts  of  a  cup.) 

76.  Find  the  radius  of  a  circle  having  the  same  area  as  the  lateral  area 
of  a  cone  of  revolution  whose  radius  is  4  and  slant  height  is  9. 

76.  The  frustum  of  a  circular  cone  is  15  in.  high  and  the  bases  are  circles 
whose  radii  are  3  in.  and  5  in.     Find  the  edge  of  an  equivalent  cube. 

77.  The  frustum  of  a  right  circular  cone  has  a  slant  height  of  9  ft. 
and  the  radii  are  5  ft.  and  7  ft.     Find  the  lateral  area  and  the  total  area. 
What  is  the  length  of  the  altitude  of  this  frustum  ?     Find  the  altitude 
of  the  cone  that  was  removed  to  leave  this  frustum.     Now  find  in  two 
ways  the  volume  of  the  frustum. 


BOOK   IX 

THE   SPHERE 

687.  A  sphere  is  a  solid  bounded  by  a  surface,  all  points 
of  which  are  equalty  distant  from  a  point  within,  called  the 
center.  The  surface  of  a  sphere  is  called  a  spherical  surface. 

A  radius  of  a  sphere  is  a 
straight  line  drawn  from  the 
center  to  any  point  of  the  surface. 

A  diameter  of  a  sphere  is  a 
straight  line  that  contains  the 
center  and  has  its  extremities  in 
the  surface. 

Ex.  1.  What  is  the  locus  of  all  points  2  in.  from  the  surface  of  a 
sphere  whose  radius  is  10  in.  ? 

Ex.  2.  Consider  the  center  of  a  sphere  at  one  of  the  vertices  of  a  rec- 
tangular box  or  room.  What  part  of  the  sphere  is  within  the  box  or 
room? 

Ex.  3.  Name  several  familiar  objects  that  are  usually  regarded  as 
spheres. 

Ex.  4.  What  is  the  shape  of  the  celestial  bodies  ? 

Ex.  5.  Why  do  you  believe  the  earth  to  be  spherical  ? 

Ex.  6.  Why  do  you  believe  the  moon  to  be  spherical  ? 

Ex.  7.  Describe  fully  the  locus  of  points  3  in.  from  a  line  8  in.  long. 


Historical  Note.  Archimedes  was  the  discoverer  of  the  formulas  for  the 
surface  and  volume  of  the  sphere.  Menelaus  (100  A.D.)  gave  us  the  proper- 
ties of  spherical  triangles. 


•V; 


394  BOOK   IX.     SOLID   GEOMETRY 

PROPOSITION  I.     THEOREM 
688.     Every  plane  section  of  a  sphere  is  a  circle. 


M 


Given :  Sphere  whose  center  is  O ;  plane  MN  intersecting 
sphere  in  AB. 

To  Prove  :  The  figure  AB  is  a  O. 

Proof :    Draw  OD  _L  to  plane  JOT,  meeting  the  plane  at  D. 
Take  P  and  Q,  any  two  points  on  the  perimeter  of  the  sec- 
tion, and  draw  DP,  DQ,  OP,  OQ. 

A  ODP  and  ODQ  are  rt.  A  (?). 

In  right  A  ODP  and  ODQ, 

OD  =  OD  (Iden.). 

OP  =  OQ  </  (687). 

.•.  the  right  A  are  congruent  (?). 

Hence                              DP  =  DQ  (?). 

That  is,  all  points  of  the  boundary  of  AB  are  equally  dis- 
tant from  D. 

Hence          the  section  AB  is  a  circle  (179).      Q.E.D. 

689.  A  great  circle  of  a  sphere  is  a  section  of  the  sphere 
made  by  a  plane  containing  the  center  of  the  sphere/ 

A  small  circle  of  a  sphere  is  a  section  of  the  sphere  made 
by  a  plane  that  does  not  contain  the  center  of  the  sphere. 

The  axis  of  a  circle  of  a  sphere  is  the  diameter  of  the 
sphere  perpendicular  to  the  plane  of  the  circle. 

The  poles  of  a  circle  of  a  sphere  are  the  ends  of  its  axis. 


THE   SPHERE  395 

A  quadrant  (in  spherical  geometry)  is  one  fourth  of  a 
great  circle. 

Equal  spheres  are  spheres  having  equal  radii.  ^ 

690.  A    plane  is  tangent  to    a   sphere   if   it   touches   the 
sphere  in  one  and  only  one  point.     Two  spheres  are  tangent 
to  each  other  if  they  are  tangent  to  the  same  plane  at  the 
same  point.     They  may  be  tangent  internally  or  externally. 

A  line  is  tangent  to  a  sphere  if  it  touches  the  sphere  in 
one  and  only  one  point  and  does  not  intersect  it. 

A  line  is  tangent  to  the  circle  of  a  sphere  if  it  lies  in  the 
plane  of  the  circle  and  touches  the  circle  in  one  and  only 
one  point.  In  all  cases  this  common  point  is  the  point  of 
contact  or  point  of  tangency. 

691.  A  sphere  is  inscribed  in  a  polyhedron  if  all  the  faces 
of  the  polyhedron  are  tangent  to  the  sphere. 

A  sphere  is  circumscribed  about  a  polyhedron  if  all  the 
vertices  of  the  polyhedron  are  in  the  spherical  surface. 

692.  The  distance  between  two  points  on  the  surface  of 
a  sphere   is  the  less  arc  of  a  great  circle  passing   through 
them. 

The  distance  between  a  point  on  a  circle  of  a  sphere  and 
the  nearer  pole  of  the  circle  is  the  polar  distance  of  the  point. 

693.  The   angle   between  two  intersecting   curves  is   the 
angle  formed  by  their  tangents  at  the  point  of  intersection. 

A  spherical  angle  is  the  angle  between  two  great  circles 
of  a  sphere. 

PRELIMINARY    THEOREMS 

694.  THEOREM.     All  radii  of  a  sphere  are  equal.  (687.) 

695.  THEOREM.     All  radii  of  equal  spheres  are  equal.  (689.) 

696.  THEOREM.     All  diameters  of  the  same  sphere  or  of 
equal  spheres  are  equal. 


396  BOOK   IX.     SOLID   GEOMETRY 

697.  THEOREM.     All  great  circles  of  the  same  sphere  or 
of  equal  spheres  are  equal. 

698.  THEOREM.     In  the  same  sphere  or  in  equal  spheres: 

I.  Equal  plane  sections  are  equally  distant  from  the  center. 

II.  Plane   sections   equally  distant   from   the   center   are 
equal.     [Converse.] 

III.  Of  two  unequal  plane  sections,  the  greater  is  at  the 
less  distance  from  the  center. 

IV.  Of   two   plane   sections   unequally   distant    from   the 
center,    the    section   at    the    greater   distance    is    the    less. 
[Converse.]     • 

In  each  case  the  diameters  of  the  sections  are  chords  of 
great  circles.     These  theorems  follow  from  208-211. 

699.  THEOREM.     Two  great  circles  of  a  sphere  bisect  each 
other.  (Because  they  have  a  common  diameter.) 

700.  THEOREM.     Every  great  circle  of  a  sphere  bisects  the 
sphere  and  the  spherical  surface.  (?)• 

701.  THEOREM.     A  sphere  may  be  generated  by  the  revo- 
lution of  a  semicircle  about  the  diameter  as  an  axis.          (?). 

702.  THEOREM.     One    and   only   one    great   circle    can  be 
drawn  through  two  points,  not  the  ends  of  a  diameter,  on  the 
surface  of  a  sphere.  (477-) 

703.  THEOREM.     One  and   only  one   circle   can  be   drawn 
through  three  points  on  the  surface  of  a  sphere.          (^77.) 

704.  THEOREM.     A  point  is  without  a  sphere  if  its  distance 
from  the  center  is  greater  than  the  radius,  and  conversely  if 
a  point  is  without  a  sphere,  its  distance  from  the  center  is 
greater  than  the  radius. 

705.  THEOREM.     The  axis  of  a  circle  of  a  sphere  passes 
through  its  center.  (This  is  proved  in  the  proof  of  688.) 


THE   SPHERE  397 

THEOREMS  AND   DEMONSTRATIONS 

PROPOSITION  II.     THEOREM 

706.    A  plane  perpendicular  to  a  radius  of  a  sphere  at  its 
extremity  is  tangent  to  the  sphere. 


Given  :  Radius  OA  of  sphere  O,  and  plane  MN  J_  to  OA  at  A. 
To  Prove  :  MN  tangent  to  the  sphere. 

Proof :  Take  any  point  X  in  MN  (except  A)  and  draw  OX. 

ox>  OA  (504,  I). 

.-.  X  lies  without  the  sphere  (704). 

Hence  every  point  of  plane  MN,  except  A,  is  without  the 

sphere  ;  that  is,  plane  MN  is  tangent  to  the  sphere.        (690.) 

Q.E.D. 

PROPOSITION  III.     THEOREM 

707.   A  plane  tangent  to  a  sphere  is  perpendicular  to  the 
radius  drawn  to  the  point  of  contact.     [Converse.] 

Given:  Plane  MN  tangent  to  sphere  O  at  A;  radius  OA. 
To  Prove  :   OA  is  _L  to  plane  MN. 

Proof :  Every  point  in  MN,  except  A,  is  without  the  sphere 

(690). 
Take  any  point  X  in  MN  and  draw  OX. 

Now  ox  is  >  OA  (704). 

That  is,  OA  is  the  shortest  line  from  O  to  MN. 

.'.  OA  is  i.  to  MN  (504,  I).      Q.E.D. 


398  BOOK   IX.     SOLID   GEOMETRY 

PROPOSITION  IV.     THEOREM 

708.  All  points  in  the  circumference  of  a  circle  of  a  sphere 
are  equally  distant  from  either  pole;  that  is,  the  polar  dis- 
tances of  all  points  in  the  circumference  of  a  circle  are  equal. 


Given :  P  and  L,  the  poles  of  O  C  on  sphere  O,  and  great 
©  PAL,  PEL. 

To  Prove  :     arc  PA  =  arc  PB  ; 
arc  AL  =  arc  BL. 
Proof  :   Draw  the  axis  PL  meeting  plane  of  O  C  at  C.    Draw 

AC  AP   BC   BP 

PC  is  _L  to  plane  DAB     (Def.  of  axis,  689). 

.-.  chord  PA  =  chord  PB  (?). 

Hence  arc  PA  =  arc  PB  (?). 

Likewise  arc  AL  ==  arc  BL.  Q.E.D. 

709.  COROLLARY.  The  polar  distance  of  a  great  circle  is  a 
quadrant. 

Ex.  1.  If  the  radius  of  a  sphere  is  26  in.  and  a  plane  is  passed,  10  in. 
from  the  center,  find  the  radius  of  the  circular  section. 

Ex.  2.  What  geographical  circles  on  the  earth's  surface  are  great 
circles  ?  Which  are  small  circles  ? 

Ex.  3.  Can  two  circles  on  the  surface  of  a  sphere  intersect  in  more 
than  two  points  ?  Why  ? 

Ex.  4.  The  area  of  a  section  of  a  sphere  45  in.  from  the  center  is  784  ir 
sq.  in.  Find  the  radius  of  the  sphere. 

Ex.  6:  The  area  of  a  section  of  a  sphere  7  in.  from  the  center  is  576  TT 
sq.  in.  Find  the  area  of  a  section  6  in.  from  the  center. 


THE   SPHERE  399 

PROPOSITION  V.     THEOREM 

710.  If  a  point  on  the  surface  of  a  sphere  is  at  the  distance 
of  a  quadrant  from  two  other  points  on  the  surface,  not  the 
ends  of  a  diameter,  it  is  the  pole  of  the  great  circle  containing 
these  two  points. 


Given  :  P,  a  point,  and  R  and  2V,  two  other  points  (not  the 
ends  of  a  diameter),  —  all  on  the  surface  of  sphere  O ;  arcs  PR 
and  P2V,  quadrants  ;  great  circle  ARNB. 

To  Prove :  P  is  the  pole  of  O  ARNB. 

Proof :   Draw  the  radii  OP,  OR,  ON. 

A  PON  and  FOR  are  rt.  A  (232). 

.-.  PO  is  _L  to  plane  AB  (485). 

.-.  PO  is  the  axis  of  O  ARNB  (Def.  689). 

.-.  P  is  the  pole  of  O  ARNB  (Def.  689). 

Q.E.D. 

Ex.  1.  Considering  the  earth  as  a  true  sphere,  what  is  the  pole  of  the 
equator?  If  a  city  has  a  latitude  of  43°,  what  is  its  polar  distance? 
What  is  the  polar  distance  of  a  place  upon  the  equator  ? 

Ex.(  2.  In  the  diagram  above,  0  is  a  trihedral  angle.  What  arcs  are 
the  measures  of  its  face  angles?  What  is  an  isosceles  trihedral  angle? 
Explain  by  this  diagram  that  the  name  is  consistent  with  the  etymology 
of  the  word. 

Ex.  3.  Prove  that  two  lines  tangent  to  a  sphere  at  a  point  determine 
a  plane  tangent  to  the  sphere  at  that  point. 

Ex.  4.  Find  the  volume  of  a  cube  circumscribed  about  a  sphere 
whose  radius  is  6  in.  Find  the  volume  of  the  cube  inscribed  in  this 
sphere. 


400  BOOK   IX.     SOLID   GEOMETRY 

PROPOSITION  VI.     THEOREM 

711.  A  spherical  angle  is  measured  by  the  arc  of  a  great 
circle  having  the  vertex  of  the  angle  as  a  pole  and  intersected 
by  the  sides  of  the  angle. 


Given :  Spherical  Z.AVB\  arc  AB  of  great  O  whose  pole  is 
F,  on  sphere  O. 
To  Prove  :  Z.AVB  is  measured  by  arc  AB. 

Proof:   Draw  radii  OJ,  O5,  OF,  and  at  F  draw  FC  tangent 

to  O  VA,  and  VD  tangent  to  O  VB.    VB  is  a  quadrant.    (709.) 

OF  is  -L  to  VD  (203), 

OF  is -L  to  0£  (232)r/ 

OF  is  -L  to  FC  and  to  OA  (?). 

.*.  FZ>  is  II  to  OB,  and  FC  is  II  to  OA  (62). 

...  ^  CVD=Z.AOB  (499). 

Z  CVD  is  the  spherical  £  AVB  (693). 

/.  AOB  is  measured  by  arc  AB  (232). 

.-.  /.  CVD  is  measured  by  arc  AB  (Ax.  6). 

That  is,  £  AVB  is  measured  by  arc  AB  (Ax.  6). 

Q.E.D. 

712.  COROLLARY.     All  arcs  of  great  circles  containing  the 
pole  of  a  great  circle  are  perpendicular  to  the  great  circle. 

713.  COROLLARY.     A  spherical  angle  is  equal  to  the  plane 
angle  of  the  dihedral  angle  formed  by  the  planes  of  its  sides. 

714.  COROLLARY.    If  two  great  circles  are  perpendicular  to 
each  other,  each  contains  the  pole  of  the  other.     (528;  689.) 


THE   SPHERE 


401 


PROPOSITION  VII.     THEOREM 
715.   A  sphere  may  be  inscribed  in  any  tetrahedron. 


Given:   Tetrahedron  A-BCD.  To  Prove  :   (?). 

Proof:  Pass  plane  OAB  bisecting  dih.  Z  AB,  and  plane 
OBC  bisecting  dih.  Z  BC,  and  plane  OCD  bisecting  dih.  Z  CD, 
the  three  planes  meeting  at  point  O. 

Point  O,  in  plane  OAB,  is  equally  distant  from  faces  ABC 
and  ABD.  (?.) 

Point  O,  in  plane  OBC,  is  equally  distant  from  faces  ABC 
and  BCD.  (?.) 

Point  O,  in  plane  OCD,  is  equally  distant  from  faces  BCD 

and  ACD.  (?.) 

.*.   O  is  equally  distant  from  all  four  faces   (Ax.  1). 

Hence  the  sphere  constructed  with  O  as  a  center  and  OB 
as  a  radius,  is  tangent  to  each  of  the  four  faces. 

.-.  that  sphere  is  inscribed  in  the  tetrahedron  (691).  Q.E.D. 

716.  COROLLARY.  The  six  planes  bisecting  the  six  dihedral 
angles  of  any  tetrahedron  meet  in  a  point. 


Ex.  1.  The  volume  of  any  polyhedron  circumscribed  about  a  sphere, 
is  equal  to  one  third  the  product  of  the  surface  of  the  polyhedron  and 
the  radius  of  the  sphere. 

PROOF  :  Pass  planes  each  containing  the  center  of  the  sphere  and  two 
vertices  of  the  polyhedron.  These  form  pyramids  whose  altitude  .  .  .  etc. 

Ex.  2.     What  is  true  of  the  point  at  which  the  six  planes  meet  in  716  ? 

Ex.  3.  Explain,  so  that  a  blind  boy  could  understand  it,  the  process  of 
inscribing  a  sphere  in  a  tetrahedron. 


402  BOOK   IX.     SOLID   GEOMETRY 

PROPOSITION  VIII.     THEOKEM 
717.    A  sphere  may  be  circumscribed  about  any  tetrahedron, 


Given :   (?).     To  Prove  :   (?). 

Proof :  Take  E  and  F,  the  centers  of  circles  circumscribed 
about  the  faces  ACD  and  BCD,  respectively.  Erect  EG  and 
FH  _L  to  these  faces.  Find  Jf,  the  midpoint  of  edge  CD. 

EG  is  the  locus  of  all  points  equally  distant  from  points 

A,  D,  and  C.  (511.) 
FH  is  the  locus  of  all  points  equally  distant  from  points 

B,  C,  and  D.  (?.) 
That  is,  all  points  in  EG  and  FH  are  equally  distant  from 

C  and  D.  (Ax.  1.) 

But  all  points  equally  distant  from  C  and  D  are  in  a  plane 
J-  to  CD  at  M.  (510.) 

.-.  EG  and  FH  are  in  this  plane  and  are  not  parallel. 

(Not  J_  to  the  same  plane.) 
That  is,          EG  and  FH  must  intersect  at  o. 
Hence  O  is  equally  distant  from  A,  J5,  C,  and  D.      (Ax.  1.) 
That  is,  using  O  as  a  center  and  OA,  or  OB,  or  OC,  or  OD, 
as  a  radius,  a  sphere  may  be  circumscribed  about  the  tetra- 
hedron A-BCD.  (691.)      Q.E.D. 

718.  COROLLARY.     Through  any  four  points  not  in  the  same 
plane  a  sphere  may  be  described. 

719.  COROLLARY.     The  six  planes  perpendicular  to  the  edges 
of  any  tetrahedron  at  their  midpoints  meet  in  a  point. 


THE   SPHERE 


403 


PROPOSITION  IX.     THEOREM 

720.  The  intersection  of  two  spherical  surfaces  is  a  circle 
whose  plane  is  perpendicular  to  the  line  which  joins  the  cen- 
ters of  the  spheres  and  whose  center  is  in  that  line. 


Given:   Two  intersecting  circles  O  and  o';  common  chord 
CD  ;  line  of  centers  XY,  intersecting  CD  at  M. 

To  Prove  :  The  spherical  surfaces  generated  by  the  revo- 
lution of  these  (D  intersect  in  a  circle. 

Proof  :  If  these  ©  are  revolved  upon  XY  as  an  axis,  they 
will  generate  spheres.  (701.) 

CM=MD  (219). 

Point  c,  common  to  both  <D,  will  generate  the  intersection 

of  the  spherical  surfaces.  (466.) 

CM  is  always  _L  to  XY     .  (219). 

.•.  the  curved  line  generated  by  C  is  in  one  plane      (486). 

.-.  the  intersection  is  a  circle  (179). 

Also  the  plane  of  this  O  is  _L  to  OOf  (486). 

And  the  center  of  the  O  is  Jf,  in  oof 


Q.E.D. 


Ex.  1.     What  can  be  said  about  the  point  mentioned  in  719  ? 

Ex.  2.  Explain,  so  that  a  blind  boy  could  understand,  how  to  circum- 
scribe a  sphere  about  a  tetrahedron. 

Ex.  3.  Find  a  point  equally  distant  from  four  points  in  space,  not  all 
in  the  same  plane. 

Ex.  4.  Practical  illustrations  of  the  truth  of  720  can  be  given  by  use 
of  soap  bubbles,  or  by  two  eggs. 

ROBBINS'S    NEW    SOLID    GEOM.  -  11 


404 


BOOK    IX.     SOLID   GEOMETRY 


CONSTRUCTION  PROBLEMS 
PROPOSITION  X.     PROBLEM 
721.     To  find  the  radius  of  a  material  sphere. 


Given  :  A  material  sphere.      Required  :   To  find  its  radius. 

Construction :  First,  place  one  point  of  the  compasses  at  P, 
and  using  any  opening  of  the  compasses,  as  AP,  with  the 
other  point  draw  a  circumference  on  the  surface  of  the 
sphere. 

Upon  this  circumference  take  three  points,  A  and  B  and  C, 
and  by  means  of  the  compasses  measure  the  straight  lines 
AB,  AC,  BC. 

Second,  construct  a  A  A'B'C',  whose  sides  equal  AB,  AC,  BC. 
Circumscribe  a  circle  about  this  A,  and  draw  the  radius  A'D'. 

Third,  construct  a  right  AP'A"D",  whose  hypotenuse 
equals  the  known  line  PA  and  whose  leg  equals  the  known 
radius  A'D'.  At  A"  erect  A"R'  -L  to  P'Aff  meeting  P'D", 
produced,  at  Rr.  Bisect  P'R'  at  o'. 

Statement :     O'P'  =  the  required  radius.  Q.E*F. 

Proof:  P  is  equally  distant  from  A,  B,  and  C         (Const.). 

D  is  equally  distant  from  A,  B,  and  C  (179). 

O  is  equally  distant  from  A,  B,  and  C  (694). 

.-.if  the  diameter  PDOR  could  be  drawn  it  would  be  _L  to 

the  plane  of  ABC. 


CONSTRUCTION   PROBLEMS  405 

.•.  Z  PDA  is  a  rt.  /.  (-173). 

Now                        A  ABC  ^  A  A'B' cr  (78). 

.'.DA  =  D'Ar  (188). 

Also                       A  PDA  3*  A  P'D"A"  (84). 


Now  Z  p^fl  is  a  rt.  Z  (240). 

Also  A  p#,4  ^  A  Pfi'^i"  (77). 

.'.PR=P'Rf  (?). 

.-.  OP=o'Pr  (Ax.  3). 

That  is,  OP=  the  radius  of  the  sphere.  Q.E.D. 

NOTE.  The  most  common  construction  on  the  surface  of  a  sphere  is 
the  drawing  of  a  great  circle.  For  this  construction,  the  compasses  must 
open  so  that  the  distance  between  their  points  is  equal  to  the  chord  of  a 
quadrant  of  a  great  circle.  To  find  this  length,  as  in  722,  we  must  find 
the  radius  of  the  solid  sphere,  as  in  721. 

PROPOSITION  XI.     PROBLEM 
722.     To  find  the  chord  of  a  quadrant  of  a  material  sphere. 


Given:   (?). 
Required:   (?). 


Construction:  Find  the  radius  of  the  sphere  (by  721). 
Using  this  radius  OP  and  any  center  O,  describe  a  semicircle 
PMR.  Erect  radius  Olf-Lto  the  diameter  P.R,  and  draw  PM. 

Statement :  Arc  PM  is  a  quadrant  of  a  great  circle  of  the 
given  sphere  and  chord  PM  is  the  required  chord.  Q.E.F. 

Proof:  Arc  PM  is  a  quadrant        (?).     Q.E.D. 


406  BOOK  IX.     SOLID   GEOMETRY 

PROPOSITION  XII.     PROBLEM 

723.     To  describe  a  great  circle  through  two  given  points 
on  the  surface  of  a  sphere. 


Given :  The  points  A  and  B  on  the  surface  of  the  sphere  O. 
Required :  To  describe  a  great  circle  through  A  and  B. 

Construction :  Find  the  chord  of  a  quadrant  of  the  given 
sphere  (by  722). 

Place  one  point  of  the  compasses  at  A,  and  using  the 
chord  just  found  as  an  opening,  describe  an  arc  on  the  sur- 
face of  the  sphere. 

Place  one  point  of  the  compasses  at  7i,  and  using  the  same 
opening,  describe  an  arc,  meeting  the  former  arc  at  P. 

Now  place  one  point  of  the  compasses  at  P  and  describe 
the  circle  BAC,  using  the  same  opening  as  before. 

Statement :   Circle  BAG  is  the  required  great  circle. 

Proof :   Points  A  and  B  are  each  at  the  distance  of  a  quad- 
rant from  P.  (Construction.) 
.•.   O  BAG  is  a  great  circle  whose  pole  is  P  (710).      Q.E.D. 


Ex.  1.     A  line  tangent  to  a  sphere  lies  in  the  plane  tangent  to  the 
sphere  at  the  same  point. 

Proof :     The  line  is  J_  to  the  radius  drawn  to  the  point  of  contact  (?). 
The  plane  also  is  _L  to  the  radius  (?). 
.-.  the  line  is  in  the  plane  (486). 

Ex.  2.     At  a  point  on  the  surface  of  a  sphere  there  can  be  only  one 
tangent  plane. 


CONSTRUCTION  PROBLEMS  407 

PROPOSITION  XIII.     PROBLEM 

724.    To  draw  an  arc  of  a  great  circle  through  a  given  point 
on  the  surface  of  a  sphere  and  perpendicular  to  a  given  great 

circle. 

P 


Given :  Point  A  on  sphere  O,  and  great  circle  BC,  whose 
pole  is  P. 

Required :  To  draw  through  A  an  arc  of  a  great  circle  _L 
to  the  great  circle  EC. 

Construction :  Place  one  point  of  the  compasses  at  A  and, 
with  an  opening  equal  to  the  chord  of  a  quadrant  of  the 
given  sphere,  describe  an  arc  of  a  great  circle  intersecting  the 
given  great  circle  at  D. 

Now  place  one  point  of  the  compasses  at  D  and  similarly 
draw  arc  of  great  circle  PAE.  Draw  PZ>,  the  arc  of  a  great 
circle  (by  723). 

Statement:  Arc  PAE  is  _L  to  circle  BEDC.  Q.E.F. 

Proof :  ED,  EP,  and  PD  are  quadrants  (709). 

.-.  E  is  the  pole  of  arc  PD 

Hence  Z.  PED  is  measured  by  quadrant  PD 

.-.  /-  PED  is  a  right  angle  (232). 

That  is,  arc  PAE  is  _L  to  circle  BEDC.  Q.E.D. 


Ex.  1.  Construct  a  plane  tangent  to  a  sphere  at  a  given  point  on  the 
surface. 

Ex.  2.  Construct  a  plane  tangent  to  a  sphere  from  a  given  point  with- 
out the  sphere.  How  many  such  planes  are  there  ? 


408  BOOK   IX.     SOLID   GEOMETRY 

SPHERICAL  TRIANGLES 

725.  A  spherical  triangle  is  a  portion  of  the  surface  of  a 
sphere  bounded  by  three  arcs  of  great  circles. 

The  bounding  arcs  are  the  sides  of  the  triangle. 

The  intersections  of  the  sides  are  the  vertices  of  the 
triangle. 

The  angles  formed  by  the  sides  are  the  angles  of  the  tri- 
angle. 

Spherical  triangles  are  equilateral,  equiangular,  isosceles, 
scalene,  acute,  right,  obtuse,  under  the  same  conditions  as  in 
plane  triangles. 

726.  A  birectangular  spherical  triangle  is  a  spherical  tri- 
angle two  of  whose  angles  are  right  angles. 


4\\: 


A 

f    /  \ 


A  trirectangular  spherical  triangle  is  a  spherical  triangle 
all  of  whose  angles  are  right  angles. 

The  unit  usually  employed  in  measuring  the  sides  of  a 
spherical  triangle  is  the  degree. 

It  is  obvious  that  three  great  circles  (not  meeting  at  a 
point)  divide  the  surface  of  a  sphere  into  eight  spherical 
triangles. 

727.  Two  spherical  triangles  are  mutually  equilateral  if  the 
sides  of  the  triangles  are  equal  each  to  each  ;  and  they  are 
mutually  equiangular  if  their  angles  are  equal  each  to  each. 


SPHERICAL   TRIANGLES 


409 


MUTUALLY  EQUILATERAL 
SPHERICAL  TRIANGLES 


POLAR 
TRIANGLES 


MUTUALLY  EQUIANGULAR 
SPHERICAL  TRIANGLES 


728.  If  three  great  circles  are  described,  having  as  their 
poles  the  vertices  of  a  spherical  triangle,  one  of  the  eight 
triangles  thus  formed  is  the  polar  triangle  of  the  first. 

The  polar  triangle  is  the  one  whose  vertices  are  nearest  the  vertices  of 
the  original  triangle. 

729.  Symmetrical  spherical  triangles  are  triangles  that  have 
their   parts   equal   but   arranged    in    reverse   order.      They 
correspond  to  symmetrical  trihedral  angles. 

Vertical   spherical  triangles   correspond   to  vertical  trihe- 
dral angles. 


If  the  diameters  of  a  sphere  are  drawn  to  the  vertices  of  a  spherical 
triangle,  the  original  triangle  and  the  triangle  whose  vertices  are  the 
opposite  ends  of  these  diameters  are  vertical  spherical  triangles. 

730.  A  spherical  polygon  is  a  portion  of  the  sphere 
bounded  by  three  or  more  arcs  of  great  circles. 

Two  spherical  polygons  are  congruent  if  they  can  be  made 
to  coincide.  The  diagonal  of  a  spherical  polygon  is  the  arc 
of  a  great  circle  connecting  two  vertices  not  in  the  same  side. 

Only  convex  spherical  polygons  are  considered  in  this  book. 


410 


BOOK   IX.     SOLID   GEOMETRY 


PRELIMINARY  THEOREMS 

731.  THEOREM.     The  planes  of  the  sides  of  a  spherical  tri- 
angle form  a  trihedral  angle: 

I.  Whose  vertex  is  the  center  of  the  sphere. 

n.  Each  of  whose  face  angles  is  measured  by  the  inter- 
cepted side  of  the  triangle.  (232.) 

III.  Each  of  whose  dihedral  angles  is  equal  to  the  corre- 
sponding angle  of  the  triangle.  (713.) 

732.  THEOREM.     Two    symmetrical   spherical    triangles  are 
mutually  equilateral  and  mutually  equiangular. 

(51,  193,  522,  713.) 

733.  THEOREM.     The  homologous  parts  of  two 
symmetrical  spherical  triangles  are  arranged  in 
reverse  order. 

Proof :  If  the  eye  is  at  the  center  of  the 
sphere,  the  order  of  the  vertices  J.,  .B,  C  is  the 
same  in  direction  as.  the  motion  of  the  hands 
of  a  clock.  But  the  order  of  A1,  B' ,  cf  is  in  the 
opposite  direction.  (See  541,  Note.)  Hence 
the  parts  are  arranged  in  reverse  order.  Q.E.D. 

734.  THEOREM.     The  homologous  parts  of  two  symmetrical 
spherical  triangles  are  equal.  (732.) 


735.  THEOREM.  Two  symmetrical 
isosceles  spherical  triangles  are  con- 
gruent. 

Proof  :  The  method  of  superposi- 
tion, as  in  the  case  of  plane  triangles. 


Historical  Note.  It  was  not  until  the  seventeenth  century  that  polar 
triangles  were  invented  by  Gerard  of  Holland.  It  was  he  also  who  found  the 
formulas  for  the  area  of  a  spherical  triangle  and  of  a  spherical  polygon. 


SPHERICAL   TRIANGLES 


411 


THEOREMS  AND   DEMONSTRATIONS 

PROPOSITION  XIV.     THEOREM 

736.    One  side  of  a  spherical  triangle  is  less  than  the  sum 
of  the  other  two. 


Given :       (?).     To  Prove  :  AB  <  AC  +  BC. 

Proof  :   Draw  radii  OA,  OB,  OC. 

In  the  trihedral  Z  o, 

Z  AOB  <  Z  AOC  +  Z  BOG 

Z  AOB  is  measured  by  arc  AB,  etc. 

.-.  arc  AB  <  arc  AC  -f-  arc  BC 

PROPOSITION  XV.     THEOREM 


(548). 

CO: 

(Ax.  6). 

Q.E.D. 


737.  In  a  birectangular  spherical  triangle  the  sides  opposite 
the  right  angles  are  quadrants,  and  the  third  angle  is  meas- 
ured by  the  third  side. 


Given  :   Birectangular  A  ABC',  Z  B  and  Z  c,  right  A. 
To  Prove  :  I.  AB  and  AC  quadrants. 

II.  Z  A  is  measured  by  arc  BC. 


412 


BOOK   IX.     SOLID   GEOMETRY 


Proof  :   I.    Draw  radii  OA,  OB,  OC. 
Arc  AB  is  -L  to  arc  BC  and 

arc  AC  is  J_  to  arc  BC  (Hyp.). 

.-.   A  is  the  pole  of  arc  BC  (714)- 

.-.  AB  and  AC  are  quadrants         (709). 
II.   Z  A  is  measured  by  arc  BC.     (711). 

Q.E.D. 


738.    COROLLARY.     The  three  sides  of  a  trirectangular  spher- 
ical triangle  are  quadrants. 


Ex.  1.     If  two  sides  of  a  spherical  triangle  are  quadrants,  the  triangle 
is  birectangular.  (710,  712.) 

Ex.  2.     If  all  sides  of  a  spherical  triangle  are  quadrants,  the  triangle 
is  trirectangular. 


PROPOSITION  XVI.     THEOREM 

739.    The  sum  of  the  sides  of  any  spherical  polygon  is  less 
than  330°. 


Given:  (?).     To  Prove  :  (?). 

Proof :   Draw  radii  to  the  several  vertices  of  the  polygon, 
forming  the  polyhedral  Z  O. 

Then      Z  AOB  +  Z  BOG  +  Z  COD  +  Z  AOD  <  360°     (549). 

But  Z  AOB  is  measured  by  arc  AB,  etc.  (?). 

.-.  arcs  AB  +  BC+CD  +  AD  <  360°     (Ax.  6).     Q.E.D. 


SPHERICAL   TRIANGLES  413 

740.  COROLLARY.     The  sum  of  the  sides  of  any  spherical 
polygon  is  less  than  the  circumference  of  a  great  circle. 

PROPOSITION  XVII.     THEOREM 

741.  If  one  spherical  triangle  is  the  polar  of  a  second  tri- 
angle, then  the  second  is  the  polar  of  the  first. 


Given  :  Spherical  A  ABC  and  its  polar  A  A'B'C!. 

To  Prove  :  A  ABC  is  the  polar  A  of  A  A'B'C'. 

Proof :  A  is  the  pole  of  arc  BrCf  (Hyp.). 

.-.  B'  is  the  distance  of  a  quadrant  from  A          (709). 
C  is  the  pole  of  arc  A'B'  (?). 

.-.  Bf  is  the  distance  of  a  quadrant  from  C  (?). 

Hence  Br  is  the  pole  .of  arc  AC  (710). 

Also      Ar  is  the  pole  of  BC,  and  (f  is  the  pole  of  AB. 

.-.  ABC  is  the  polar  A  of  A  A'B'C'  (728). 

Q.E.D. 

NOTE.  Many  properties  of  a  trihedral  angle  are  common  to  the  cor- 
responding spherical  triangle.  The  polyhedral  angle  is  similarly  related 
to  the  spherical  polygon.  Sometimes  it  is  advantageous  to  employ  one, 
sometimes  the  other.  The  spherical  triangle  is  perhaps  simpler  and 
more  suggestive  of  properties  than  the  corresponding  trihedral  angle, 
when  the  plane  angles  of  its  dihedral  angles  appear  in  the  same  diagram. 

It  is  a  most  instructive  and  helpful  exercise  for  the  student  to  draw 
spherical  triangles  and  their  polars,  etc.,  on  a  material  sphere,  such  as  a 
slate  globe,  a  large  apple,  a  ball,  or  other  spherical  object.  A  great  many 
geometrical  truths  can  be  fixed  in  the  mind  by  an  orange  and  three  long 
needles. 


414  BOOK  IX.     SOLID   GEOMETRY 

PROPOSITION  XVIII.     THEOREM 

742.     In  two  polar  spherical  triangles  each  angle  of  one  and 
the  opposite  side  of  the  other  are  supplementary. 


Given :  Polar  A  ABC  and  A'B'C'. 

To  Prove:  Z  A  +  a'  =  180° ;  Zvl'  +  a 

Z  V  +  V  =  180°;  Z  B'  +  5  =  180°; 
Zc+<?'  =  180°;  Zc'+<?=180°. 

Proof:   Prolong  arc  ^'c'  to  meet  arc  4J3  at  R  and  arc  ^1C 

at  S'                        n'8  =  90°  and  c'R  =  90°  (709). 

.'.B'S+C'B=18Q°  (Ax.  2). 

That  is,      C'S  +  B'C"  +  C'R  or  RS  +  7?'e'  =  180°  (Ax.  4). 

Now                   US  is  the  measure  of  Z  A  (711)- 

Also  2?'C'  =  a'. 

.-.  Z^  +  a'  =  1800  (Ax.  6). 
Similarly,     Z  7?  +  6'  =  180°  ;  Z  c  +  <*  =  180°. 

Again,  prolong  arcs  AfBr  and  ^4'c'  to  meet  arc  BC  at  Jf 

and  L'  BL  =  90°  and  CM  =  90°  (709). 

.-.  BL  +  CM=1SQ°  (Ax.  2). 

That  is,  LM+  MB+CM=  180°,  or  LJf  +  BC  =  180° 

(Ax.  4). 

Now  LM  is  the  measure  of  Z.  A!  (?). 

Also  BC  =  a. 

.-.  Z^l'  +  «=1800  (Ax.  6). 

Similarly,     ^  Bf  +  6=180°;  Z  c'+ <?  =  180°.     Q.E.D. 


SPHERICAL   TRIANGLES  415 

743.  COROLLARY.      In   two   polar   spherical  triangles   each 
angle  of  one  is  measured  by  the  supplement  of  the  opposite 
side  of  the  other. 

PROPOSITION  XIX.     THEOREM 

744.  The  sum  of  the  angles  of    a  spherical  triangle  is 
greater  than  180°  and  less  than  540°. 


Given  :  A  spherical  A  ABC. 

To  Prove  :     I.     Z.A  +  Z.U  +  Z.  c  >  180°  ; 
II.     Z.A  +  Z  B  +  Z.  c  <  540°. 

Proof  :   I.     Construct  A  A'H'C',  the  polar  A  of  A  ABC. 

^A  +  a'  =  180°,  Z  B  +  br  =  180°,  Z  C  +  c'  =  180°        (742). 

Adding,      ^A  +  Z.B  +  ^.C  +  a1  +  &  +  €>  =  540°     (Ax.  2). 

But 

Subtracting, 


V  +  b'  +  c' 

<  360°        (739). 

Z^+Z^+ZC 

>  180°     (Ax.  9). 

Q.E.D. 

'.A  +  ZB  +  Z.C+a'+b'  +  c' 

=  540°     (Ax.  2). 

a'  +  V  +  <?' 

>  0°           (725). 

But 

Subtracting,         Z^  +  Z7i  +  ZC  <  540°    (Ax.  9). 

Q.E.D. 

746.     COROLLARY.     The  sum  of  the  angles  of  a  spherical  tri- 
angle is  greater  than  two,  and  less  than  six,  right  angles. 

746.     COROLLARY.     A  spherical  triangle  may  have  one,  two, 
or  three  obtuse  angles. 


416 


BOOK   IX.     SOLID   GEOMETRY 


PROPOSITION  XX.     THEOREM 
747.     Two  symmetrical  spherical  triangles  are  equal. 


B 


Given  :  Two  symmetrical  spherical  A  ABC  and  A'B'C'. 
To  Prove  :  A  ABC  =  A  A'B'C'. 

Proof  :  Suppose  P  is  the  pole  of  the  O  containing  A,  B,  C. 
Draw  the  diameters  AAf,  BB',  cc',  PP\  and  the  arcs  of  great 
©,  PA,  PB,  PC,  P'A',  P'B',  P'C'. 


(193). 

(708). 
(Ax.  1). 

(735). 


.-.  arc  PA  =  arc  P'A' 

Also  arc  PB  =  arc  P'B'  and  arc  PC=  arc  P'C'. 
But  PA  =  PB  =  PC 

.-.  P'A'  =  P'B'  =  P'C' 
Hence  A  APB  ^  A  A'  P'B' 

A  AGP  =*  A 
A  ^PC  ^  A 
Adding, 


A  ABC  =  A 


'  (Ax.  2)- 

Q.E.D. 

NOTE.  If  the  pole  P  should  be  without  the  A  ABC,  one  of  the  pairs 
of  equal  A  would  be  without  the  original  &  and  would  be  subtracted 
from  the  sum  of  the  others  to  obtain  &  ABC  and  A'B'C'. 

748.  COROLLARY.  Vertical  spherical  triangles  are  symmet- 
rical and  equal. 

Ex.  1.     Are  symmetrical  spherical  triangles  ever  congruent? 
Ex.  2.     What  unit  is  used  in  measuring  the  sides  and  angles  of  a 
spherical  triangle? 


SPHERICAL  TRIANGLES  417 

PROPOSITION  XXI.     THEOREM 

749.  Provided  two  spherical  triangles  on  the  same  sphere 
(or  on  equal  spheres)  have  their  parts  arranged  in  the  same 
order,  they  are  congruent : 

I.  If  two  sides  and  the  included  angle  of  one  are  equal 
respectively  to  two  sides  and  the  included  angle  of  the  other. 

II.  If  a  side  and  the  two  angles  adjoining  it  of  one  are 
equal  respectively  to  a  side  and  the  two  angles  adjoining  it  in 
the  other. 

III.  If  three  sides  of  the  one  are  equal  respectively  to 
three    sides    of    the    other;    that   is,  if    they  are  mutually 
equilateral. 


On  Equal  Spheres. 


Given:   (?). 

To  Prove  :  A  ABC  ^  A  RST. 

Proof :  I  and  II.     Superposition  as  in  plane  A. 
III.     Draw  radii  of  the  sphere  to  all  the  vertices  of  the  A. 
The  face  A  of  the  trih.  Z  O  =  the  face  A  of  the  trih.  Z  N, 
respectively.  (?.) 

Hence  trih.  Z  o  =  trih.  Z  N  (546). 

.-.  dih.  Z  OA  =  dih.  Z  NR  ;  dih.  /.OB  —  dih.  Z  N8 ;  etc. 

.-.  the  A  are  mutually  equiangular  (731,  III). 

Hence  the  A  can  be  made  to  coincide. 

.'.  A  ABC  ^  A  RST  (26). 

Q.E.D. 


418  BOOK   IX.     SOLID   GEOMETRY 

PROPOSITION  XXII.     THEOREM 

750.  Provided  two  spherical  triangles  on  the  same  sphere 
(or  on  equal  spheres)  have  their  parts  arranged  in  reverse 
order,  they  are  symmetrical : 

I.  If  two  sides  and  the  included  angle  of  one  are  equal 
respectively  to  two  sides  and  the  included  angle  of  the  other. 

II.  If  a  side  and  the  two  angles  adjoining  it  of  one  are  equal 
respectively  to  a  side  and  the  two  angles  adjoining  it  of  the 
other. 

III.  If  three  sides  of  one  are  equal  respectively  to  three  sides 
of  the  other;  that  is,  if  they  are  mutually  equilateral. 


Given :   (?).    To  Prove :   (?). 

Proof :  In  each  of  these  cases  construct  a  third  spherical 
A  R'S'T*,  symmetrical  to  the  A  RST. 

Then  AR'S'T'  will  have  its  parts  equal  to  the  parts  of 
A  ABC  and  arranged  in  the  same  order. 

.-.  AR'S'T'^AABC  (749). 

Hence  A  RST  is   symmetrical  to  A  ABC  (Ax.  6). 

Q  E.D 

751.  COROLLARY.  Two  mutually  equilateral  spherical  tri- 
angles are  mutually  equiangular  and  are  congruent  or  sym- 
metrical. 

When  are  they  congruent?     When  are  they  symmetrical? 


Ex.     Is  the  corresponding  theorem  about  plane  triangles  true? 


SPHERICAL   TRIANGLES  419 

PROPOSITION  XXIII.     THEOREM 

752.  Two  mutually  equiangular  spherical  triangles  on  the 
same  sphere  (or  on  equal  spheres)  are  mutually  equilateral, 
and  are  congruent  or  symmetrical. 


Given  :  A  A  and  A1,  mutually  equiangular. 

To  Prove  :  A  A  and  Ar  mutually  equilateral,  and  congruent 
or  symmetrical. 

Proof :  Construct  A  E  and  E1,  the  polar  A  of  A  and  A'. 
The  sides  of  E  are  supplements  of  the  A  of  A. 


(742V 
The  sides  of  E1  are  supplements  of  theziof  A' .  j 

But  the  A  of  A  are  =  respectively  to  the  A  of  Af     (Hyp.). 
.-.  &E  and  E1  are  mutually  equilateral  (49). 

Hence  A  E  and  Ef  are  mutually  equiangular  (751). 

Again  A  A  and  A1  are  the  polar  A  of  E  and  Ef  (741). 

.•.  the  sides  of  A  are  supplements  of  the  A  of  E  )  ,?, 

the  sides  of  A'  are  supplements  of  the  A  of  E'  \ 
Hence         A  A  and  A1  are  mutually  equilateral  (?). 

.  *.  they  are  congruent  (when  ?) ; 

or  symmetrical  (when  ?). 

Q.E.D. 

Ex.  Ts  the  corresponding  theorem  about  plane  triangles  true  ?  Is 
there  any  theorem  concerning  the  congruence  of  plane  triangles  that  is 
not  true  of  spherical  triangles?  If  the  parts  of  two  plane  triangles  were 
arranged  in  reverse  order  and  they  were  kept  in  a  plane,  could  they  be 
made  to  coincide  ? 

ROBBINS'S    NEW    SOLID    GEOM. 12 


420 


BOOK   IX.     SOLID   GEOMETRY 


PROPOSITION  XXIV.     THEOREM 

753.   The  angles  opposite  the  equal  sides  of  an  isosceles 
spherical  triangle  are  equal. 


Given:   (?). 

To  Prove  :  Z  B  =  Z  C. 


Proof :   Suppose  X  the  midpoint  of  BC. 

Draw  AX,  the  arc  of  a  great  Q.     Now  the  two  spherical 
A  ABX  and  AXC  are  mutually  equilateral.     (Explain.) 
.-.  they  are  mutually  equiangular  and  symmetrical     (751). 
.-.  Z£  =  ZC  (734).     Q.E.D. 

754.  COROLLARY.     The  arc  of  a  great  circle  drawn  from  the 
vertex  of  an  isosceles  spherical  triangle  to  the  midpoint  of  the 
base  bisects  the  vertex  angle  and  is  perpendicular  to  the  base. 

PROPOSITION  XXV.     THEOREM 

755.  If  two  angles  of  a   spherical  triangle  are  equal,  the 
sides  opposite  are  equal. 


Given:  (?).     To  Prove:  (?). 


SPHERICAL   TRIANGLES 


421 


Proof :   Construct  A  A'B'C',  the  polar  A  of  A  ABC. 
Then        A' B1  is  the  supplement  of  Z  C. 
And  A'C'  is  the  supplement  of  Z  73. 

.-.  Z7/=Z  C' 

Again  A  ABC  is  the  polar  A  of  A  A'B'C' 

AB  is  the  supplement  of  Z  (/,  and  ^C  of  Z 
.-.  ^7i  =  4C 

PROPOSITION  XXVI.     THEOREM 


(49). 
(753). 
(741). 

CO- 

(49). 

Q.E.D. 


756.  If  two  angles  of  a  spherical  triangle  are  unequal,  the 
sides  opposite  are  unequal  and  the  greater  side  is  opposite  the 
greater  angle. 


Given:  A  ABC',  Z  ABC  >  Z  C. 

To  Prove  :  AC  >  AB. 

Proof :   Suppose  BR  drawn,  the  arc  of  a  great  Q,  making 
Z  CBR  =  Z  C  and  meeting  AC  at  R. 

Now  ^72  +  7*£>47*  (736). 

But  BR  =  CR  (755). 

.-.  ^#4-  CR  >  .47?  (Ax.  6). 

That  is,  AC  >  AB  Q.E.D. 


Ex.     Compare  Propositions  XXIV,  XXV,  XXVI  with  the  correspond- 
ing theorems  about  plane  triangles. 


422  BOOK   IX.     SOLID   GEOMETRY 

PROPOSITION  XXVII.     THEOREM 

757.  If  two  sides  of  a  spherical  triangle  are  unequal,  the 
angles  opposite  are  unequal  and  the  greater  angle  is  opposite 
the  greater  side.  [Converse.] 

Given:  (?).     To  Prove :  Z  ABC  >  Zc. 
Proof  :  Z  ABC  is  either  <  Z.  C  or 

=  Z  C  or  >  Z  C. 
Continue  by  method  of  exclusion    (90). 

PROPOSITION  XXVIII.     THEOREM 

768.  If  two  circles  on  a  sphere  contain  a  point  on  the  arc  of 
a  great  circle  that  joins  their  poles,  they  have  no  other  point 
hi  common. 


Given :   Point  P  on  the  arc  AB  of  a  great  O  of  a  sphere, 
and  P  common  to  two  circles  whose  poles  are  A  and  B. 

To  Prove :  P  is  the  only  point  common  to  these  ©. 

Proof  :  Suppose  X  is  another  common  point. 
Draw  arcs  of  great  ®  AX  and  EX. 

Then                          AX-}-  BX  >  AP  +  J5P  (736). 

But                                     AX  =  AP  (708). 

Subtracting,                       BX  >  BP  (Ax.  7). 
That  is,  X  is  without  the  O  B  and  cannot  be  in  both  the 

circles.  Q.E.D. 


SPHERICAL  TRIANGLES  428 

PROPOSITION  XXIX.     THEOREM 

759.  The  shortest  line  that  can  be  drawn  on  the  surface  of 
a  sphere,  between  two  points  on  the  surface,  is  the  less  arc  of 
the  great  circle  containing  the  two  points. 


Given :  Points  A  and  B,  and  AB  the  arc  of  a  great  O  join- 
ing them ;  line  ADEB,  any  other  line  on  the  surface  of  the 
sphere,  between  A  and  B. 

To  Prove  :  Arc  AB  <  line  ADEB. 

Proof :  Take  on  arc  AB  any  point  C,  and  describe  two  cir- 
cles through  C,  having  A  and  B  as  their  poles,  and  intersect- 
ing ADEB  at  D  and  E.  Point  C  is  the  only  point  common  to 
these  two  ®.  (758.) 

No  matter  what  kind  of  line  AD  is,  a  line  of  equal  length 
can  be  drawn  from  A  to  C,  on  the  surface ;  and  a  line  can  be 
drawn  from  B  to  C  equal  in  length  to  BE. 

[Imagine  AD  revolved  on  the  surface  of  the  sphere,  using  A  as  a  pivot, 
and  D  will  move  along  the  O  to  point  C.  Similarly  with  BE.~\ 

There  is  now  a  line  from  A  to  B,  through  C,  <  ADEB. 

That  is,  whatever  the  nature  of  ADEB,  there  is  a  shorter 
line  from  A  to  B,  which  contains  C,  any  point  of  arc  AB. 

Thus  the  shortest  line  contains  all  the  points  of  AB  and 
therefore  is  the  line  AB.  Q.E.D. 

NOTE.  This  theorem  justifies  the  definition  of  the  "distance"  be- 
tween two  points,  etc.,  in  692. 


424  BOOK  IX.     SOLID  GEOMETRY 

ORIGINAL  EXERCISES 

1.  Vertical  spherical  angles  are  equal. 

2.  If  two  spherical  triangles,  on  the  same  or  equal  spheres,  are  mutu- 
ally equilateral,  their  polar  triangles  are  mutually  equiangular. 

3.  The  polar  triangle  of  an  isosceles  spherical  triangle  is  isosceles. 

4.  The  polar  triangle  of  a  birectangular  spherical  triangle  is  birec- 
tangular. 

6.   If  two  dihedral  angles  of  a  trihedral  angle  are  equal,  the  opposite 
face  angles  also  are  equal. 

Proof:  Construct  a  sphere  having  the  vertex  as  center,  etc. 

6.  If  two  face  angles  of  a  trihedral  angle  are  equal,  the  opposite  di- 
hedral angles  also  are  equal. 

7.  A  trirectangular  spherical  triangle  is  its  own  polar  triangle, 

8.  Two  symmetrical  spherical  polygons  are  equal. 

9.  Any  side  of  a  spherical  polygon  is  less  than  the  sum  of  the  other 
sides.     [Draw  diagonals  from  a  vertex.] 

10.  If  the  three  face  angles  of  a  trihedral  angle  are  equal,  the  three 
dihedral  angles  also  are  equal. 

11.  State  and  prove  the  converse  of  No.  10. 

12.  A  straight  line  cannot  meet  a  spherical  surface  in  more  than  two 
points. 

13.  If  two  dihedral  angles  of  a  trihedral  angle  are  unequal,  the  oppo- 
site face  angles  are  unequal,  and  the  greater  face  angle  is  opposite  the 
greater  dihedral  angle. 

14.  State  and  prove  the  converse  of  No.  13. 

15.  All  the  tangent  lines  drawn  to  a  sphere  from  an  external  point 
are  equal. 

16.  The  volume  of  any  tetrahedron  is  equal  to  one  third  the  product 
of  its  total  surface  by  the  radius  of  the  inscribed  sphere. 

17.  Every  point  of  a  great  circle  that  is  perpendicular  to  an  arc  at  its 
midpoint  is  equally  distant  from  the  ends  of  the  arc.  A 

18.  The  points  of  contact  of   all   lines  tangent 
to  a  sphere  from  an  external  point  lie  in  the  cir- 
cumference of  a  circle. 

19.  Any  point  in  the  arc  of  a  great  circle  that       X. 
bisects  a  spherical  angle  is  equally  distant  from  ( 

the  sides  of  the  angle.  B  ""T  C 


ORIGINAL   EXERCISES  425 

20.  If  the  opposite  sides  of  a  spherical  quadrilateral  are  equal,  the 
opposite  angles  are  equal. 

21.  If  the  opposite  sides  of  a  spherical  quadrilateral  are  equal,  the 
diagonals  bisect  each  other. 

22.  If  the  diagonals  of  a  spherical  quadrilateral  bisect  each  other,  the 
opposite  sides  are  equal. 

23.  The  exterior  angle  of  a  spherical  triangle  is  less  than  the  sum  of 
the  opposite  interior  angles. 

24.  The  sum  of  the  angles  of  a  spherical  quadrilateral  is  more  than 
four  right  angles. 

25.  If  two  spheres  are  tangent  to  each  other,  the  straight  line  joining 
their  centers  passes  through  the  point  of  contact. 

26.  The  sum  of  the  angles  of  a  spherical  polygon  is  more  than  2  n  —  4 
right  angles  and  less  than  2  n  right  angles. 

27.  The  arcs  of  great  circles   bisecting  the  angles  of  a  spherical  tri- 
angle meet  in  a  point. 

28.  If  a  tangent  line  and  a  secant  are  drawn  to  a  sphere  from  an  ex- 
ternal point,  the   tangent  is   a  mean  proportional   between   the  whole 
secant  and  the  external  segment. 

29.  The  product  of  any  secant  that  can  be  drawn  to  a  sphere  from  an 
external  point,  by  its  external  segment,  is  constant  for  all  secants  drawn 
through  the  same  point. 

30.  If  two  spherical  surfaces  intersect  and  a  plane  is  passed  contain- 
ing their  intersection,  tangents  from  any  point  in  this  plane  to  the  two 
spherical  surfaces  are  equal. 

31.  Find  the  distance  from  the  center  of  a  sphere  whose  radius  is 
15  in.  to  the  plane  of  a  small  circle  whose  radius  is  8  in. 

32.  The  polar  distance  of  a  small  circle  is  60°  arid  the  radius  of  the 
sphere  is  12  in.     Find  the  radius  of  the  circle. 

33.  The  total  surface  of  a  tetrahedron  is  90  sq.  m.,  and  the  radius  of 
the  inscribed  sphere  is  4  m.     Find  the  volume  of  the  tetrahedron. 

34.  Find  the  radius  of  the  sphere  inscribed  in  a  tetrahedron  whose 
volume  is  250  cu.  in.  and  total  surface  is  150  sq.  in. 

35.  Find  the  total  surface  of  a  tetrahedron  whose  volume  is  320  cu.  in., 
if  the  radius  of  the  inscribed  sphere  is  8  in. 


426  BOOK    IX.     SOLID   GEOMETKY 

36.  Find  the  radius  of  the  sphere  inscribed  in  a  regular  tetrahedron 
whose  edges  are  each  10  in. 

37.  Find   the  radius  of   the   sphere   circumscribed    about   a   regular 
tetrahedron  whose  edges  are  each  18  in. 

38.  Find  the  radii  of  the  spheres  inscribed  in  and  circumscribed  about 
a  cube  whose  edges  are  each  10  in. 

39.  The  sides  of   a  spherical   triangle  are  60°,  80°,  110°.     Find   the 
angles  of  its  polar  triangle. 

40.  The  angles  of  a  spherical  triangle  are  74°,  119°,  87°.     Find  the 
sides  of  its  polar  triangle. 

41.  The  chord  of  the  polar  distance  of  the  circle  of  a  sphere  is  12  m., 
and  the  radius  of  the  sphere  is  9  m.     Find  the  radius  of  the  circle. 

42.  The  polar  distance  of  a  circle  is  60°  and  the  diameter  of  the  circle 
is  8  ft.     Find  the  diameter  of  the  sphere. 

[Denote  by  R,  each  side  of  an  equilateral  triangle  whose  altitude  is  4  ft.] 

43.  The  radii  of  two  spherical  surfaces  are  11  in.  and  13  in.,  and  their 
centers  are  20  in.  apart.     Find  the  radius  of  the  circle  of  their  intersec- 
tion.    Find  also  the   distances  from   the  centers  of  the  spheres  to  the 
center  of  this  circle. 

44.  The  radii  of  two  spherical  surfaces  are  20  m.  and  37  m.,  and  the 
distance  between  their  centers  is  19m.     What  is  the  length  of  the  diame- 
ter of  their  intersection  ? 

45.  Bisect  an  arc  of  a  great  circle. 

46.  Draw  an  arc  of  a  great  circle  perpendicular  to  a  given  arc  of  a 
great  circle  through  a  given  point  in  the  arc. 

47.  Bisect  a  spherical  angle. 

48.  Bisect  an  arc  of  a  small  circle. 

49.  Circumscribe  a  circle  about  a  given  spherical  triangle. 

60.  Construct  a  spherical  angle  equal  to  a  given  spherical  angle  at  a 
given  point  on  the  same  sphere. 

61.  Construct  a  spherical  triangle  having  the  three  sides  given. 

62.  Construct  a  spherical  triangle  having  the  three  angles  given. 

63.  Construct  a  plane  tangent  to  a  sphere  at  a  given  point  on  the 
surface. 

54.    Construct  a  spherical  surface  having  the  radius  given  and  contain- 
ing three  given  points. 


SPHERICAL   AREAS   AND   VOLUMES 


427 


65.  Construct  a  spherical  surface  that  shall  have  a  given  radius,  touch 
a  given  plane,  and  contain  two  given  points. 

66.  Construct  a   spherical   surface   that   shall  have   a  given  radius, 
shall  be  tangent  to  a  given  sphere,  and  contain  two  given  points. 

57.  Construct  a  spherical  surface  that  shall  contain  four  given  points. 

58.  Construct  a  plane  that  shall  contain  a  given  line  and  be  tangent 
to  a  given  sphere. 

69.    Construct  a  plane  tangent  to  a  given  sphere  and  parallel  to  a 
given  plane. 

60.  What  is  the  locus  of  points  on  the  surface  of  a  sphere  : 

(a)  Equally  distant  from  two  given  points  on  the  surface? 

(b)  Equally  distant  from  two  given  points  not  on  the  surface  ? 

61.  What  is  the  locus  of  the  centers  of  those  spherical  surfaces  that 
pass  through  two  given  points  ? 

62.  What  is  the  locus  of  the  centers  of  the  spherical  surfaces  of  given 
radius  that  contain  two  given  points  ? 

63.  What  is  the  locus  of  the  centers  of  the  spherical  surfaces  that  pass 
through  three  given  points  ? 


SPHERICAL  AREAS  AND   VOLUMES 

760.  A  lune  is  a  portion  of  the  surface  of  a  sphere 
bounded  by  two  great  semicircles. 

The  points  of  intersection  of  the  sides  of  a  lune  are  the 
vertices  of  the  lune. 

The  angles  made  at  the  vertices  by  the  sides  are  the 
angles  of  the  lune. 


LUNE 


(a)  SPHERICAL 

SECTOR 
(6)  SPHERICAL 
PYRAMID 


ZONE 


SPHERICAL 
CONE 


SPHERICAL 
SEGMENT 


428  BOOK   IX.     SOLID   GEOMETRY 

761.  A  zone  is  a  portion  of  the  surface  of  a  sphere 
bounded  by  two  circles  whose  planes  are  parallel. 

The  bases  of  a  zone  are  the  circles  bounding  it. 

The  altitude  of  a  zone  is  the  perpendicular  distance  be- 
tween the  planes  of  its  bases. 

If  one  of  the  planes  is  tangent  to  the  sphere,  the  zone  is  a 
zone  of  one  base. 


762.  A  spherical  degree  is  y^  of  the  surface  of  a  sphere. 
If  the  surface  of  a  sphere  is  divided  into  720  equal  parts, 
each  part  is  a  spherical  degree. 

The  size  of  a  spherical  degree  depends  on  the  size  of  the  sphere. 

It  may  be  easily  conceived  to  be  half  a  lune  whose  angle  is  1  degree, 
that  is,  a  birectangular  spherical  triangle  whose  third  angle  is  lc. 

How  many  spherical  degrees  are  there  in  a  trirectangular  spherical 
triangle  ? 

763.  The  spherical  excess  of   a  spherical   triangle  is   the 
sum  of  its  angles  less  180°.     That  is,  E=  A+  B+  C—  180°. 

764.  A  spherical  pyramid  is  a  portion  of  a  sphere  bounded 
by  a  spherical  polygon  and  the  planes  of  its  sides. 

The  vertex  of  a  spherical  pyramid  is  the  center  of  the 
sphere. 

The  base  of  a  spherical  pyramid  is  the  spherical  polygon. 

765.  A  spherical  sector  is  the  solid  generated  by  the  revo- 
lution of  the  sector  of  a  circle  about  any  diameter  of  the 
circle  as  an  axis. 

The  base  of  the  spherical  sector  is  the  zone  generated  by 
the  arc  of  the  circular  sector. 

A  spherical  cone  is  a  spherical  sector  whose  base  is  a  zone 
of  one  base. 

766.  A    spherical   segment  is   a  portion  of   a   sphere    in- 
cluded   between    two    parallel    planes    that    intersect    the 
sphere. 


SPHERICAL   AREAS  AND  VOLUMES  429 

The  bases  of  a  spherical  segment  are  the  circular  sections 
made  by  the  parallel  planes. 

The  altitude  of  a  spherical  segment  is  the  perpendicular 
distance  between  the  bases. 

A  spherical  segment  of  one  base  is  a  segment  one  of 
whose  bounding  planes  is  tangent  to  the  sphere. 

A  hemisphere  is  a  spherical  segment  of  one  base,  which 
base  is  a  great  circle. 

A  spherical  wedge  is  a  portion  of  a  sphere  bounded  by  a 
lune  and  the  planes  of  its  sides. 


Ex.  1.  What  is  the  spherical  excess  of  a  spherical  triangle  whose 
angles  are  60°,  70°,  and  100°? 

Ex.  2.     Distinguish  between  a  zone  and  a  spherical  segment. 

Ex.  3.  Find  the  area  of  a  spherical  degree  on  a  sphere  whose  surface 
is  3600  sq.in. 

Ex.  4.  Find  the  area  of  a  spherical  triangle  containing  80  spherical 
degrees,  on  a  sphere  whose  surface  is  450  sq.  ft. 

Ex.  6.  Find  the  area  of  a  spherical  polygon  containing  152  spherical 
degrees  on  a  sphere  whose  surface  is  630  sq.  yd. 

Ex.  6.  A  spherical  triangle  containing  128  spherical  degrees  has  an 
area  of  72  sq.  in.  What  is  the  area  of  the  spherical  surface  ? 

PRELIMINARY  THEOREMS 

767.  THEOREM.    Either  angle  of  a  lune  is  measured  by  the 
arc  of  a  great  circle  described  with  the  vertex  of  the  lune  as  a 
pole,  and  included  between  the  sides  of  the  lune.  (711.) 

768.  THEOREM.    The  angles  of  a  lune  are  equal. 

769.  THEOREM.    Every  great  circle  of  a  sphere  divides  the 
sphere  into  two  equal  hemispheres,  and  the  surface  into  two 
equal  zones. 

770.  THEOREM.    The  spherical  excess  of  a  spherical  n-gon 

is  equal  to  the  sum  of  its  angles  less  (n  —  2)  180°. 

Proof:  (?). 


430  BOOK  IX.     SOLID   GEOMETRY 

771.  THEOREM.    If  a  regular  polygon  having  an  even  num- 
ber of  sides  is  inscribed  in,  or  circumscribed  about,  a  circle, 
and  the  figure  is  made  to  revolve  about  one  of  the  longest 
diagonals  of  the  polygon,  the  surface  generated  by  the  perim- 
eter of  the  polygon  approaches  the  surface  of  the  sphere  gene- 
rated by  the  circle,  as  a  limit,  if  the  number  of  sides  of  the 
polygon  is  indefinitely  increased. 

772.  THEOREM.    If  a  polyhedron  is  circumscribed  about  a 
sphere  and  the  number  of  its  faces  is  indefinitely  increased, 
the  surface  of  the  polyhedron  approaches  the  surface  of  the 
sphere  as  a  limit,  and  the  volume  of  the  polyhedron  approaches 
the  volume  of  the  sphere  as  a  limit. 


NOTE.  If  a  regular  polygon  having  an  even  number  of  sides  is  in- 
scribed in,  or  circumscribed  about,  a  circle,  and  the  figure  is  made  to 
revolve  about  one  of  the  longest  diagonals  of  the  polygon,  the  surface 
generated  by  the  polygon  is  composed  of  the  surfaces  of  cones,  a  cylin- 
der, and  frustums,  and  the  surface  generated  by  the  circle  is  a  spherical 
surface. 

Ex.  1.  Find  the  spherical  excess  of  a  polygon  whose  angles  are  80°, 
110°,  140°,  130°,  160°. 

Ex.  2.  The  spherical  excess  of  a  spherical  polygon  is  the  difference 
between  the  sum  of  its  angles  and  the  sum  of  the  angles  of  a  plane  poly- 
gon having  the  same  number  of  sides. 

Ex.  3.  The  sum  of  the  angles  of  a  spherical  quadrilateral  is  less 
than  eight  right  angles. 

Ex.  4.  Find  the  spherical  excess  of  a  spherical  hexagon  if  each  of  its 
angles  equals  128°.  If  each  angle  equals  155°,  find  the  excess. 

Ex.  6.  If  the  opposite  angles  of  a  spherical  quadrilateral  are  equal, 
the  opposite  sides  are  also  equal. 

Proof:  Prolong  one  pair  of  opposite  sides  in  both  directions  until  they 
meet.  Now  prove  two  triangles  congruent. 


AREAS   OF   SPHERES  431 

THEOREMS  AND   DEMONSTRATIONS 
PROPOSITION  XXX.     THEOREM 

773.  The  area  of  the  surface  generated  by  a  straight  line 
revolving  about  an  axis  in  its  plane  is  equal  to  the  product  of 
the  projection  of  the  line  upon  the  axis  by  the  circumference 
of  a  circle  whose  radius  is  the  line  perpendicular, to  the  revolv- 
ing line  at  its  midpoint,  and  terminating  in  the  axis. 

Given :  Line  AB  revolving  about 
axis  XX' ;  CD  =  projection  of  AB  on 
XX1 ';  MP  =  a  =  _L  erected  at  mid- 
point of  AB  and  terminating  in  XX1 ; 
MO  =  radius  of  mid-section. 

To  Prove :  Surface  generated  by 
AB  =  CD  •  2  Tra. 

Proof  :  I.  The  surface  generated  by  A B  is  the  surface  of 
the  frustum  of  a  right  circular  cone  whose  bases  are  gene- 
rated by  AC  and  BD,  and  the  mid-section,  by  MO. 

Area  of  surface  =  2  wMO  -  AB  (681). 

Now             A  ABH  and  MPO  are  similar  (310). 

.'.  MO  :  AH  =  MP  :  AB  (?). 

Hence                       MO  •  AB  =  AH  •  MP  =  CD  •  a  .          (?). 

.  •.  area  of  surface  =  2  TTCD  •  a  =  CD  •  2  tra  (Ax.  6). 

II.  If  AB  is  II  to  XX* r,  the  surface  is  cylindrical  and  equals 
CD  -  2  Tra  (654). 

III.  If  AB  meets  XX1  at  C,  the  entire  surface  is  conical 
and  equals  TrBD  -  AB  (680). 

Now  BD=2MO  (136). 

And  MO  -  AB  =  CD.  •  a.  (?.) 

. '.  TrBD  •  AB  =  TT  •  2  MO  -  AB  =  TT  •  2  •  CD  -  a  =  CD  •  2  Tra 

(Ax.  6). 
.-.  the  area  of  the  surface  =  CD  •  2  Tra      (Ax.  6).     Q.E.D. 


432 


BOOK   IX.     SOLID   GEOMETRY 


PROPOSITION  XXXI.     THEOREM 

774.  The  surface  of  a  sphere  is  equal  in  area  to  four  great 
circles ;  that  is,  to 


Given :    Semicircle    ACF ;     diameter   AF;    S  =  surface   of 
sphere  generated  by  revolving  the  semicircle  about  AF  as  an 
axis ;  R  =  radius  of  this  sphere. 
To  Prove  :   S  =  4  irR2. 

Proof:   Inscribe  in  this  semicircle  half  of  a  regular  poly- 
gon having  an  even  number  of  sides.     Draw  the  apothems,  a. 
Draw  the  projections  of  the  sides  of  the  polygon  on  the  diam- 
eter.    Now,  if  the  figure  revolves  on  AF  as  an  axis, 
the  surface  AB  =  AP  •  2  IT  a 

the  surface  BC  =  PS  -  2  ira  (773). 

the  surface  CD  =  8T  •  2  ira    etc.     Adding, 


the  entire  surface  =  (AP  -f-  PS+  ST  +  etc.) 
=  AF  •  2  TTO. 


jra      (Ax.  2). 
(Ax.  6). 

Now,  if  the  number  of  sides  of  the  polygon  is  indefinitely 
increased,  the  entire  surface  generated  by  the  polygon  ap- 
proaches S  as  a  limit.  (771.) 
a  approaches  R  as  a  limit                     (422). 
Also                     AF  •  2  TTQ  approaches  AF  •  2  irR. 

c  —     A  v     9  ™-K>  /^99CU 

•    .    o   —    ./I  /*    •   j-i  7T  K  \jUtiij  j  • 

But  AF=2R  (?). 

(Ax.   6).       Q.E.D. 


AREAS   OF   SPHERES  433 

775.  COROLLARY.     The  area  of  a  spherical   degree   equals 

4irR2  =  T*^*  sq.  units. 
720        180 

776.  COROLLARY.     The  areas  of  the  surfaces  of  two  spheres 
are  to  each  other  as  the  squares  of  their  radii  and  as  the 
squares  of  their  diameters. 

Proof.  s  - 

'   S' 

PROPOSITION  XXXII.     THEOREM 

777.  The  area  of  a  zone  is  equal  to  the  product  of  its  alti- 
tude by  the  circumference  of  a  great  circle. 

Given  :   (The  same  as  in  774). 

To  Prove  :  The  area  of  the  zone  generated  by  the  arc  EC  = 
PS  x  2  TTR. 

Proof:  The  area  generated  by  chord  BC=  PS-  2  ira   (773). 

If  the  number  of  sides  of  the  inscribed  polygon  is  indefi- 
nitely increased,  the  length  of  chord  BC  approaches  arc  BC 
and  the  surface  generated  by  chord  BC  approaches  the  area  of 
a  zone. 

Also  PS  -  2  ira  will  approach  PS  •  2  TTR. 

Hence     Area  of  zone  BC  =  PS    2  irB         (229).       Q.E.D. 

778.  COROLLARY.    Area  of  a  zone, 

Z  =  2 


(Where  Z  =  area  of  the  zone,  H=  its  altitude,  and  R  =  radius 
of  sphere.) 

Ex.  1.     On  a  sphere  whose  radius  is  6  in.,  find  the  area  of  a  zone  2J  in. 
in  height. 

Ex.  2.     What  does  the  formula  for  the  area  of  a  zone  become  when  the 
altitude  is  the  diameter  ?  when  the  altitude  is  half  the  radius? 


434 


BOOK   IX.     SOLID   GEOMETRY 


779.  COROLLARY.  The  area  of  a  zone  of  one  base  is  equal  to 
the  area  of  a  circle  whose  radius  is  the  chord  of  the  generating 
arc. 

xl 
A 

D 
0 


X' 

Given :   Arc  AB  of  semicircle  ABC;  diameter  AC;  chord  AB. 
To  Prove  :   Area  of  zone  generated  by  arc  AB  =  TrAB2. 


Proof :  Area  of  zone  AB  =  AD  •  2  TTR 

That  is,  area  of  zone  AB  =  TT  •  AD  •  2  B. 

Draw  chord  BC.     A  ABC  is  a  rt.  A 

.'.  AD-  AC=AB2 

That  is,  AD-2E  =  AB2 

Hence  area  of  zone  41?  =  TrAB2 

That  is,  area  of  zone  of  one  base  =  IT  (chord)2. 


(777). 


(333). 
(Ax.  6). 
(Ax.  6). 

Q.E.D. 


Ex.  1.  What  is  the  area  of  a  zone  of  one  base  whose  chord  is  7  in.  in 
length?  of  one  whose  chord  is  14  in.  in  length? 

Ex.  2.  What  does  the  formula  for  the  area  of  a  zone  of  one  base  be- 
come when  the  generating  arc  is  a  semicircle  ?  when  the  generating  arc 
is  a  quadrant? 

Ex.  3.  If  the  radius  of  the  earth  is  approximately  4000  mi.  and  the 
altitude  of  the  north  temperate  zone  is  2080  mi.,  what  is  the  area  of  the 
north  temperate  zone  ? 

Ex.  4.  Prove  that  on  the  same  or  equal  spheres,  zones  having  equal 
altitudes  have  equal  areas. 


AREAS   OF   SPHERES 


435 


PROPOSITION  XXXIII.     THEOREM 

780.  The  area  of  a  lune  is  to  the  area  of  the  surface  of  its 
sphere  as  the  angle  of  the  lune  is  to  360°. 


Given :  Lune  ABCDA  on  sphere  o ;  L  =  area  of  lune  ;  8  = 
area  of  sphere ;  great  O  EB  whose  pole  is  A. 

To  Prove  :  L  :  8  =  Z  A  :  360. 

Proof:  I.    If  arc  BD  and  the  circumference  of  O  .E^B'are 

commensurable.     There  exists  a  common   unit  of  measure. 

Suppose  this  unit  contained  5  times  in  BD;  32  times  in  the 

circumference.     .-.  arc  BD  :  circumference  =  5  :  32  (?). 

Arc  BD  measures  Z  A  (711). 

.-.  ZA  :  360  =  5:  32  (Ax.  6). 

Pass  great  CD  through  the  several  points  of  division  of 
circumference  EB  and  vertex  A,  dividing  the  surface  of  the 
sphere  into  32  equal  lunes.  Then  L  :  S  =  5  :  32  (Ax.  3). 

Hence  L  :  s  = /.  A  :  360  (Ax.  1).     Q.E.D. 

II.    If  the  arc  and  circumference  are  incommensurable. 

The  proof  is  similar  to  that  found  in  293,  524. 

781.  COROLLARY.  The  number  of  spherical  degrees  in  the 
area  of  a  lune  is  double  the  number  of  degrees  in  its  angle. 

Proof:  Let  L°  denote  the  area  of  the  lune,  expressed  in 
spherical  degrees. 

Then                L°  :  720°  =  Z  A  :  360°  (Subst.  in  780). 

.-.  Jy°  =  2Z4.  Q.E.D. 

ROBBINS'S    NEW    SOLID   GEOM. — 13 


436 


BOOK   IX.     SOLID   GEOMETRY 


782.   COROLLARY.     The  area  of  a  lune  expressed  in  square 
units  is  — 


Q.E.D. 


90 
Proof  :   Substituting  in  780,  L  :  4  TTR*  =  /.  A  :  360. 

L  =  ^A- 
90 

783.  COROLLARY.     Two  lunes  on  the  same  or  equal  spheres 
are  to  each  other  as  their  angles. 

Proof :  L  :  S  =  Z  A  :  360,  and  Lr  :  S  =  Z  A1  :  360       (780). 
Dividing,  L  :  Lf  =  Z  A  :  Z  A1  (Ax.  3).     Q.E.D. 

PROPOSITION  XXXIV.     THEOREM 

784.  The  number  of  spherical  degrees  in  a  spherical  tri- 
angle is  equal  to  the  spherical  excess  of  the  triangle. 


Lune, 


Given :   Spherical  A  ABC  on  sphere 
O ;  spherical  excess  of  the  A  =  E. 

To  Prove  :  Number  of  spherical  de- 
grees in  A  ABC  =  E. 


Proof :   Continue  the  sides  of  the  A  ABC  to  form  the  lunes 
ABA'cA,  BAB'CB,  CAC'BC\  draw  diameters  AAf,  BBf,  CCf. 
AABC'  =  AA'B'C  (747). 

Lune  CAC'BC  =  A  ABC+A  AC'B  =  A  ABC+AA'B'C  (Ax.  6). 


AREAS   OF   SPHERES  437 


Now 

AABC  +  AA'B'c=lune  CAC'BC.   } 

And 

AABC  + 

A  A'  BC  =  lune  ABA'CA. 

And 

AABC  + 

A  AB'C  =  lune  BAB'CB.  . 

(Ax.  4.) 
Adding, 

A  AB'C 
=  lune  A  +  lune  B  +  lune  C         (Ax.  2). 

Now,  first,  4  of  these  A  compose  a  hemisphere  and  =  360 
spherical  degrees ; 

second,  the  3  lunes  =  2Z^  +  2Z£  +  2ZC         (781). 

By  substituting  in  the  long  equation  above, 

2  A  ABC  +  360  =  2Z4  +  2Z£  +  2ZC     (Ax.  6). 

.-.  AABC=  Z4+Z£+Z  c-180   (Ax.  3). 

That  is,  A  ABC  =  E  (763).     Q.E.D. 

785.   COROLLARY.    The  area  of  a  spherical  triangle  expressed 
in  square  units  is —  • 

Proof :       1  spherical  degree  =  ^—  sq.  units 

180 


•.  Area  of  a  spherical  A  =  E    *R    sq.  units    (Ax.  3). 

180 

Q.E.D. 


Ex.  1.     How  many  spherical  degrees  are  there  in  a  lune  whose  angle 
equals  25°? 

Ex.  2.     On  a  sphere  whose  radius  is  10  in.,  is  a  lune  whose  angle  is  10°. 
Find  the  area  of  the  lune  in  square  inches.. 

Ex.  3.     Reduce  the  formula  for  the  area  of  a  lune  in  square  units,  if 
the  angle  is  90°;  if  the  angle  is  180°. 

Ex.  4.     On  a  sphere  whose  radius  is  9  ft.  is  a  spherical  triangle  whose 
angles  are  70°,  145°,  and  60°.     Find  the  area  of  the  triangle. 


438 


BOOK   IX.     SOLID   GEOMETRY 


PROPOSITION  XXXV.     THEOREM 

786.     The  number  of  spherical  degrees  in  a  spherical  poly- 
gon is  equal  to  its  spherical  excess. 


Given  :   A  spherical  n-gon. 

To  Prove :  The  number  of  spherical  degrees  in  this  n-gon 
=  the  excess  of  the  polygon. 

Proof :  From  any  vertex  draw  diagonals,  dividing  the 
polygon  into  (n  —  2)  A ;  let  the  sums  of  the  A  of  these  A  be 
denoted  by  s,  sv  «2,  •••  etc. 

Now  the  number  of  spherical  degrees  in  one  A  =  s  —  180° 

(784). 

Number  of  spherical  degrees  in  another  A  =  81  —  180°  (?). 

Etc.,  for  (n-2)  A. 
Adding,  the  number  of  spherical  degrees  in  the  n-gon 

=  the  sum  of  its  A-(n-  2)  180°   (Ax.  2). 
The  excess  of  w-gon  =  sum  of  its  A  -  (n  —  2)  180°    (770). 

.-.  the  number  of  spherical  degrees  in  a  spherical  polygon 
=  the  excess  of  the  polygon  (Ax.  1).  Q.E.D. 


Ex.  1.     Find  the  area  of  a  spherical  triangle  whose  angles  are  80°, 
125°,  and  95°,  on  a  sphere  whose  radius  is  6.3  in. 

Ex.  2.     Find  the  area  of  a  spherical  polygon  whose  angles  are  135°, 
105°,  85°,  155°,  120°,  on  a  sphere  whose  radius  is  15  ft. 

Ex.  3.     Find  the  area  of  a  spherical  triangle  whose  angles  are  72°, 
97°,  and  101°,  on  a  sphere  whose  radius  is  3£  in. 


VOLUMES   OF   SPHERES  439 

PROPOSITION  XXXVI.     THEOREM 
4-rrJJ8 


787.     The  volume  of  a  sphere  = 


3 


Given  :   Sphere  o ;    radius  =  R  ;    surface  =  S ;   volume  =  V. 

To  Prove:   F  =  i^-3. 
3 

Proof :  Suppose  a  polyhedron  circumscribed  about  the 
sphere,  its  surface  denoted  by  Sf,  and  its  volume  by  V1. 

Suppose  planes  are  passed  through  the  edges  of  the  poly- 
hedron and  the  center  of  the  sphere,  thus  dividing  the  poly- 
hedron into  pyramids  whose  vertices  are  all  at  the  center, 
and  whose  common  altitude  is  B. 

The  volume  of  one  such  pyramid  =  ^  E  •  its  base        (612). 

.*.  volume  of  all  pyramids  =  ^  E  •  the  sum  of  all  the  bases 

(Ax.  2). 
That  is,  V'  =  IR.8'. 

Indefinitely  increase  the  number  of  faces  of  the  polyhe- 
dron, thus  indefinitely  decreasing  each  face, 

and  v'  approaches  V  as  a  limit  1  f  772^ 

and  s'  approaches  S  as  a  limit  I 
Hence        £  R  •  8f  approaches  J  R  •  8  as  a  limit. 

.-.  V=%R-  8  (229). 

But  S  =  47TE2  (?). 

(Ax.  6).     Q.E.D. 


440  BOOK   IX.     SOLID   GEOMETRY 

788.  COROLLARY.    The  volumes  of  two  spheres  are  to  each 
other  as  the  cubes  of  their  radii  or  as  the  cubes  of  their 
diameters. 

Proof: 

F    _  4  7T.R3        4  TTR'3 R3    _   ( J  _D)3  __  D3  ,  *  n^ 

V1  ~~       3  3       ~  R'B       (i  D')3      D'3 

Q.E.D. 

789.  COROLLARY.     The  volume  of  a   spherical  pyramid  is 
equal  to  one  third  the  product  of  the  polygon  that  is  its  base, 
by  the  radius  of  the  sphere. 

F=  I  (area  of  base)U. 
Proof :  Similar  to  the  proof  of  787. 

790.  COROLLARY.    The  volume  of  a  spherical  wedge  is  to  the 
volume  of  the  sphere  as  the  angle  of  its  base  is  to  360°. 

Proof  :   Similar  to  the  proof  of  780. 

791.  COROLLARY.    Volume  of  a  spherical  wedge, 


270 

(Where  A  =  the  ^  of  the  lune,  and  R  =  the  radius  of  sphere.) 
Pr.oof:  F  :     wiz8  =  Z      :  360°  (790). 


270 

792.  COROLLARY.    The  volume  of  a  spherical  sector  is  equal 
to  one  third  the  product  of  the  zone  that  is  its  base  by  the 
radius  of  the  sphere. 

Proof  :  Similar  to  the  proof  of  787. 

793.  COROLLARY.     Volume  of  a  spherical  sector  or  a  spher- 
ical cone,  r=iz.*=|irl*«ir. 

(Where  F=the  volume  of    the    spherical    sector    or    cone, 
H  =  the  altitude  of  its  base,  JB  =  the  radius  of  the  sphere.) 


VOLUMES   OF   SPHERES 


441 


PROPOSITION  XXXVII.     PROBLEM 

794.   To  derive  a  formula  for  the  volume  of  a  spherical 
segment. 

There  are  Three   Cases 

1.    Spherical  segment  of  one  base. 

x  x 

^*-**^  ^ 

A, 


/  Spherical  Cone        \  (  Con e          .  ^ ^  I  —-  f-SpherrcdltieS'nent     I 

(Generated  by  OAXJ   ~~  I  Generated  by  A  CO  I  —  (Generated  byACXJ 


Given  :  Spherical  segment  generated  by  the  figure  AGX\ 
semicircle  XAY  ';  AG=r\>  radius  of  sphere  =  #;  altitude  = 
CX  =  H. 

Required  :  To  find  the  volume  of  the  spherical  segment. 

Computation  :   Draw  chords  AX,  AY,  and  radius  AO. 
The  right  A  AGO  will  generate  a  cone  of  revolution 

(Def.  660). 

The  volume  of  spherical  segment  ACX  =  the  volume  of 
spherical  cone  OAX  minus  the  volume  of  cone  AGO. 


(793). 
(683). 
(331,  II)  . 

-  #)  (Ax.  6). 

Hence  volume  of  spherical  segment 

ACX  =  |  7TR2ri  —  (|  7TR2H  —  7TRII2  +  J  7TH3)       (AX.    6). 

.  •.  Volume  of  spherical  segment  of  one  base  =  ^  irlT2  (  3  H  —  H)  . 

Q.E.F. 


Volume  of  spherical  cone  OAX  =  f  TrR2  -  H 
Volume  of  cone  AGO  =  %  Trr2  -  CO 
Now  r2  =  CX  •  CY  =  J/(2  B  -  H) 

Also  CO  =  R  —  H. 

.-.  vol.  AGO  = 


442 


BOOK   IX.     SOLID   GEOMETRY 


2.    Spherical  segment  not  including  the  center. 
A  ^ -^  A. 


tor    I   i    /  O?/7e 

Generated  by  ABOJ    I    I  Generated  by ACO 


Generated  by  BDO)  —  (Generated  by  ACDB  / 


Given :  Spherical  segment  generated  by  figure       . 
ACDB ;  semicircle  XABY;    AC  —  r\    BD  =  rr;    ra-    gy 
dius  of  sphere  =  12 ;  altitude  =  CD  =  H. 

Required  :  To  find  the  volume,  F,  of  the  sphc:  - 
ical  segment. 

Computation:  The   A  AGO   and   BDO   generate 
cones  of  revolution  (Def.  660). 

Denote  OD  by  d. 

The  volume  of  spherical  segment  ACDB 

=  the  volume  of  spherical  sector  ABO 
plus  the  volume  of  cone  AGO 
minus  the  volume  of  cone  BDO. 

Now  the  volume  of  spherical  sector  ABO  =  %  7rR2H      (792). 

And  the  volume  of  cone  AGO  =  J  irr\d  +  H)  (683). 

And  the  volume  of  cone  BDO  =  4  7rr'2d  (683). 


F= 


-f[* 


-r'2)] 

But  in  rt.  A  AGO,  R2  =  r2  +  (d  +  tf)2 
and  in  rt.  A  BDO,       R2  =  r'2  +  d2  .     . 
Subtracting  and  solving, 

d  =  —— — — 

Substituting  in  (3), 

~4    I     *.'4     i      rr4  9 

R2  =  T   t-  r     f  a. 


(1) 

(2)1 

(3)1 


(4) 


(334) 


ORIGINAL   EXERCISES  443 

Substituting  (4)  and  (5)  in  (1),  and  simplifying, 

3 


_ 

= 


rrzH  +  fl3"! 
2  J 


.-.  V  =    TrH(r*  +  r'*)  +    TrH*.  Q.E.F. 

3.    Spherical  segment  including  the  center. 
Given  :   Spherical  segment  generated  by  figure  BDSR  ;  etc. 
Required  :  To  find  the  volume,  F,  of  the  spherical  segment. 
Computation:    v  =  the  volume  of  spherical  sector  BOB 
plus  the  volume  of  cone  BDO 
plus  the  volume  of  cone  R80. 
(Computation  similar  to  that  in  2,  with  same  final  formula.) 

ORIGINAL   EXERCISES 

1.  Prove  that  the   area  of   the  surface  of  a  sphere  is  equal  to  the 
square  of  the  diameter  multiplied  by  TT  ;  that  is,  5  =  TrZ)2. 

2.  Prove  that  the  volume  of  a  sphere  is  equal  to  one  sixth  the  cube 
of  the  diameter  multiplied  by  TT  ;  that  is,  V  =  $  TrZ)8. 

3.  The  surface  of  a  sphere  is  equal  to  the  cylindrical  surface  of  the 
circumscribed  cylinder. 

4.  The  total  surface  of  a  hemisphere  is  three  fourths  the  surface  of 
the  sphere. 

6.    The  'volume  of  a  sphere  is  two  thirds  the  volume  of  the  circum- 
scribed cylinder. 

6.  Upon  the  same  circle  as  a  base  are  constructed  a  hemisphere,  a 
cylinder  of   revolution,  and   a  cone  of  revolution,  all  having  the  same 
altitude.     Prove  that   their  total  areas  are   3  TrR2,  4  Tr#2,  TrZZ2(l  +  \/2), 
respectively,  and  their  volumes  are  f  wRs,  TrR3,  j  TrR8,  respectively. 

7.  Two  zones  on  the  same  sphere,  or  on  equal  spheres,  are  to  each 
other  as  their  altitudes. 

8.  The  area  of  the  surface  of  a  sphere  is  equal  to  the  area  of  the 
circle  whose  radius  is  the  diameter  of  the  sphere. 

9.  Show  that  the  formula  for  the  volume  of  a  spherical  segment  of 
one  base  reduces  to  the  correct  formula  for  the  volume  of  a  hemisphere 
when  the  base  of  the  segment  is  a  great  circle  ;  and  to  the  correct  for- 
mula for  the  volume  of  a  sphere  when  the  planes  are  both  tangent. 


444 


BOOK    IX.     SOLID    GEOMETRY 


a  circle  whose   radius  is   R,  there  are   in- 
square   and  an  equilateral   triangle   having 


10.  In  an  equilateral  triangle  is  inscribed  a  circle, 
and  the  figure  is  revolved  about  an  altitude  of  the  tri- 
angle as  an  axis.     Prove  : 

(a)  That  the  surface  generated  by  the  circumfer- 
ence is  two  thirds  the  lateral  surface  generated  by  the 
triangle. 

(&)  That  the  volume  generated  by  the  circle  is  four 
ninths  the  volume  generated  by  the  triangle. 

11.  Derive  a  formula  for  the  surface  of  a  sphere,  containing  only  V 
and  TT. 

12.  Derive  a  formula  for  the  volume  of  a  sphere,  containing  only  S 
and  TT. 

13.  In 
scribed  a 

their  bases  parallel ;  the  whole  figure  is  then  revolved 
about  the  diameter  perpendicular  to  the  base  of  the 
triangle.  Find,  in  terms  of  R : 

(a)  The  total  areas  of  the  three  surfaces  generated. 

(6)  The  volumes  of  the  three  solids  generated. 

14.  If  a  cylinder  of  revolution  having  its  altitude  equal  to  the  diam- 
eter of  its  base,  and  a  cone  of  revolution  having  its  slant  height  equal  to 
the  diameter  of  its  base,  are  both  inscribed  in  a  sphere : 

(a)  The  total  area  of  the  cylinder  is  a  mean  proportional  between 
the  area  of  the  surface  of  the  sphere  and  the  total  area  of  the  cone. 

(6)  The  volume  of  the  cylinder  is  a  mean  proportional  between  the 
volume  of  the  sphere  and  the  volume  of  the  cone. 

16.    About  a  circle  whose  radius  is  a  there  are  circumscribed  a  square 
and  an  equilateral  triangle  having  their  bases  in  the 
same  straight  line.    The  whole  figure  is  then  revolved 
about  an  altitude  of  the  triangle.    Find,  in  terms  of  a  : 

(a)  The  total  areas  of  the  three  surfaces   gen- 
erated. 

(&)  The  volumes  of  the  three  surfaces  generated. 

16.  If  a  cylinder  of  revolution  having  its  altitude  equal  to  the  diam- 
eter of  its  base,  and  a  cone  of  revolution  having  its  slant  height  equal  to 
the  diameter  of  its  base,  is  circumscribed  about  a  sphere : 

(a)  The  total  area  of  the  cylinder  is  a  mean  proportional  between  the 
area  of  the  surface  of  the  sphere  and  the  total  area  of  the  cone. 

(7>)  The  volume  of  the  cylinder  is  a  mean  proportional  between  the 
volume  of  the  sphere  and  the  volume  of  the  cone. 


r-. 

A 


ORIGINAL   EXERCISES  445 

17.  The  line  joining  the  centers  of  two  intersecting  spherical  surfaces 
is  perpendicular  to  the  plane  of  the  intersection  at   the  center  of   the 
intersection. 

18.  A  cube  and  a  sphere  have  equal  surfaces.     Show  that  the  sphere 
has  the  greater  volume. 

19.  Prove  that  the  parallel  of    latitude  through  a  point  having  30° 
north  latitude  bisects  the  surface  of  the  northern  hemisphere. 

20.  Prove  that  in  order  that  the  eye  may  observe 
one  sixth  of  the  surface  of  a  sphere  it  must  be  at  a 
distance  from  the  center  of  the  sphere  equal  to  f  of 
the  radius. 

Proof:  Zone  TT  =  $  surface  of  sphere  (Hyp.). 

.-.  AB  =  $  diain.  =  |  R.     Hence  BC  -  f  R. 

In  rt.  A  ETC,  TC2  =  EC-  BC  (?)  ;  .-.  R*  =  EC  •  f  R,  or  EC  =  f  R. 
(Explain.)  Q.E.D. 

21.  How  many  miles   above-  the  surface  of  the  earth  (diameter  of 
earth  =  7960  mi.)  must  a  person  be  in  order  that  he  may  see  one  sixth 
of  the  earth's  surface  ? 

22.  If  the  area  of  a  zone  of  one  base  is  a  mean  proportional  between 
the  area  of  the  remaining  zone  of  the  sphere  and  the  area  of  the  entire 
sphere,  the  altitude  of  the  zone  is  R(Vo  —  1). 

23.  The  area  of  a  lune  is  to  the   area  of  a  trirectangular  spherical 
triangle  as  the.  angle  of  the  lune  is  to  45°. 

24.  A  cone,  a  sphere,  and  a  cylinder  have  the  same  diameters  and 
altitudes.     Prove  that  their  volumes  are  in  arithmetical  progression. 

26.  The  surface  of  a  sphere  bears  the  same  ratio  to  the  total  surface 
of  the  circumscribed  cylinder  of  revolution  as  the  volume  of  the  sphere 
bears  to  the  volume  of  the  cylinder. 

26.  The  smallest  circle  upon  a  sphere  whose  plane  passes  through  a 
given  point  within  the  sphere,  is  the  circle  whose  plane  is  perpendicular 
to  the  diameter  through  the  given  point. 

27.  What  part  of  the  surface  of  the  earth  could  one  see  if  he  were  at 
the  distance  of  a  diameter  above  the  surface? 

28.  Prove  that  if  any  number  of  lines  in  space  are  drawn  through  a 
point,  and  from  any  other  point  perpendiculars  to  these  lines  are  drawn, 
the  feet  of  all  of  these  perpendiculars  lie  on  the  surface  of  a  sphere. 

29.  The  volume  of  a  sphere  is  to  the  volume  of  the  circumscribed 
cube  as  TT  :  6.     The  volume  of  a  sphere  is  to  the  volume  of  the  inscribed 
cube  as  TT  :  f  \/3. 


446  BOOK   IX.     SOLID   GEOMETRY 

30.  There  are  five  spheres  that  touch  the  four  planes  of  the  faces  of 
a  tetrahedron. 

31.  If  two  angles  of  a  spherical  triangle  are  supplementary,  the  sides 
of  the  polar  triangle,  opposite  these  angles,  are  supplementary. 

32.  A  square,  whose  side  is  a,  is  revolved  about  a  diagonal,  and  also 
about  an  axis  bisecting  two  opposite  sides.     Which  of  these  figures  con- 
tains the  greater  volume?     Which  has  the  greater  surface? 

33.  Find  the  area  of  the  surface  and  the  volume  of  a  sphere  whose 
radius  is  6  in. 

34.  Find  the  area  of  a  zone  whose  altitude  is  4  in.  on  a  sphere  whose 
radius  is  14  in. 

35.  Find  the  area  of  a  lune  whose  angle  is  30°  on  a  sphere  whose  radius 
is  8  in. 

36.  Find  the  area  of  a  spherical  triangle  whose  angles  are  110°,  41°, 
92°  on  a  sphere  whose  radius  is  10  in. 

37.  Find  the  volume  of  a  sphere  whose  radius  is  5  m. 

38.  Find  the  volume  of  a  spherical  pyramid  whose  base  is  35  sq.  in. 
on  a  sphere  whose  radius  is  12  in. 

39.  Find  the  area  of  a  spherical  polygon  whose  angles  are  87°,  108°, 
121°,  128°  on.  a  sphere  whose  radius  is  25  cm. 

40.  What  is  the  radius  of  a  sphere  whose  surface  is  1386  sq.  yd.  ? 


41.  What  is  the  radius  of  a  sphere  whose  volume  is  --       cu.  in.  ? 

3 

42.  What  is  the-  area  of  the  surface  of  a  sphere  whose  volume  is 
2887rcu.  ft.? 

43.  What  is  the  volume  of  a  sphere  the  area  of  whose  surface  is 
2464  sq.  in.  ? 

44.  Find  the  area  of  a  zone  whose  altitude  is  3£  in.  if  the  radius  of 
the  sphere  is  7£  in. 

46.    Find  the  volume  of  a  spherical  sector  the  altitude  of  whose  base 
is  5£  in.  if  the  radius  of  the  sphere  is  6  in. 

46.  Find  the  diameter,  the  circumference  of  a  great  circle,  and  the 
volume  of  a  sphere  the  area  of  whose  surface  is  25  TT  sq.  ft. 

47.  By  how  many  cubic  inches  is  a  9-in.  cube  greater  than  a  9-in. 
sphere  ? 

48.  The  radius  of  a  sphere  is  15  in.,  and  the  angles  of  the  base  of  a 
spherical  pyramid  are  160°,  127°,  96°,  145°,  and  117-.     Find  the  volume 
of  the  pyramid. 


ORIGINAL   EXERCISES 


447 


49.  A  cylindrical  vessel  10  in.  in  diameter  contains  a  liquid.  A  metal 
hall  is  immersed  in  the  liquid  and  the  surface  rises  £  in.  What  is  the 
diameter  of  the  ball? 

60.  If  a  sphere  3  ft.  in  diameter  weighs  99  lb.,  how  much  will  a  sphere 
of  the  same  material  4  ft.  in  diameter  weigh  ? 

61.  The  radii  of  the  bases  of  a  frustum  of  a  cone  of  revolution  are 
5  in.  and  6  in.,  and  the  altitude  of  the  frustum  is  19|  in.     What  is  the 
diameter  of  an  equal  sphere  ? 

62.  What  is  the  radius  of  a  sphere  whose  surface  is  equal  to  the  total 
surface  of  a  right  circular  cylinder  having  an  altitude  equal  to  21  in.  and 
radius  of  the  base  equal  to  6  in.  ? 

53.  Find  the  volume  generated  by  the  revolution  of  an  equilateral 
triangle  inscribed  in  a  circle  whose  radius  is  8  in.  about  an  altitude  of 
the  triangle  as  an  axis.  (See  Fig.  of  Ex.  55.) 

64.  In  the  figure  of  Ex.  55,  find  the  volume  of  the 
segment  generated  by  the  figure  A  ED  revolving  about 
CD  as  an  axis. 

65.  Find  the  area  of  the  surface  and  the  volume 
of  the  sphere  generated  by  a  circle  that  is  circum- 
scribed about  an  equilateral  triangle  whose  side  is  10  in. 

66.  Circumscribing  a  sphere  whose  radius  is  18  m.  is 
a  cylinder  of  revolution.     Compare  their  total  areas ; 
their  volumes. 

67.  Circumscribing  a  cylinder  of  revolution  whose 
altitude  and  diameter  are  each  6  in.  is  a  sphere.     Find 
the  volume  and  area  of  the  surface  of  the  sphere. 

58.  Circumscribing  a  cylinder  whose  altitude  is  4  in. 
and  diameter  is  3  in.  is  a  sphere.     Find  the  radius  and 
volume  of  the  sphere. 

59.  Each  edge  of  a  cube  is  8  in.     What  is  the  area  of  the  surface 
and  the  volume  of  the  circumscribed  sphere  ? 

60.  Find  the  volume  of  one  of  the  segments  cut  from  a  10  in.  sphere 
by  the  plane  of  one  of  the  faces  of  the  inscribed  cube. 

61.  The  volume  of  a  certain  sphere  is  179f  cu.  ft.     Find  the  radius 
of  a  sphere  8  times  as  large.     Find  the  radius  of  a  sphere  3  times  as 
large. 

62.  The  radius  of  a  certain  sphere  is  5  in.     What  is  the  radius  of  a 
sphere  twice  as  great  ?  half  as  great  ?  two  thirds  as  great  ? 


448  BOOK   IX.     SOLID   GEOMETRY 

63.  A  hollow  sphere  lias  an  outer  diameter  of  20  in.  and  an  inner 
diameter  of  10  in.     Find  the  volume  of  the  metal  in  the  shell. 

64.  Find  the  diameter  of  that  sphere  whose  volume  is,  numerically, 
equal  to  the  area  of  its  surface. 

66.  A  projectile  consists  of  a  right  circular  cylinder  having  a  hemi- 
sphere at  each  end.  If  the  cylinder  is  9  in.  long  and  7  in.  in  diameter, 
what  is  the  volume  of  one  projectile  ? 

66.  Inscribed  in  a  regular  tetrahedron  whose  edge  is  4  in.,  and  circum- 
scribed about  it,  are  two  spheres.     Find  their  radii. 

67.  Find   the   radii  of   the  spheres  inscribed    in   and   circumscribed 
about  a  regular  hexahedron  whose  edge  is  8  m. 

68.  Find   the  radii  of   the  spheres    inscribed   in   and   circumscribed 
about  a  regular  octahedron  whose  edge  is  12  in. 

69.  How  many  spherical  bullets  \  in.  in  diameter  can  be  made  from 
a  cube  of  lead  5  in.  on  each  edge? 

70.  The  area  of  a  spherical  triangle  whose  angles  are  158°,  77°,  95° 
is  288|  sq-  ft.     Find  the  radius  of  the  sphere. 

71.  The   area  of  a  spherical  triangle  whose   excess  is  75°  is  135  IT 
sq.  in.     Find  the  radius  of  the  sphere. 

72.  If  the  radius  of  a  sphere  is  2.5  in.,  and  the  sides  of  a  triangle  on 
it  are  104°,  115°,  101°,  find  the  area  of  the  polar  triangle. 

73.  In  a  trihedral  angle  the  plane  angles  of  the  dihedral  angles  are 
75°,   85°,   110°.     Find  the   number  of   degrees  of   surface  of  a   sphere 
whose  center  is  the  vertex  of  the  trihedral  angle  inclosed  by  the  faces 
of  this  trihedral  angle. 

74.  What  is  the  area  of  a  spherical  hexagon,  each  of  whose  angles  is 
145°,  on  a  sphere  whose  radius  is  15  m.  ? 

75.  How  many  miles  above  the  earth  must  a  person  be  in  order  that 
he  may  see  a  third  of  its  surface?  one  eighth  of  its  surface? 

76.  Find  the  altitude  of  the  zone  whose  area  is  equal  to  the  area  of 
a  great  circle  of  a  sphere. 

77.  If  the  radius  of  a  sphere  is  doubled,  how  is  the  amount  of  surface 
affected?  the  volume?  the  weight? 

78.  At  a  distance  (=  d)  from  the  center  of  a  sphere  whose  radius  is  r 
is  an  illuminating  point.     What  is  the  altitude  of  the  zone  illuminated? 

79.  On  a  sphere  having  a  radius  of  5  in.  is  an  equiangular  spherical 
triangle  whose  area  is  5  ir  sq.  in.     Find  the  angles  of  the  triangle. 

80.  Find  the  area  of  the  surface  of  a  sphere  whose  volume  is  1  cu.  yd. 


ORIGINAL   EXERCISES  449 

81.  Find  the  volume  of  a  sphere  whose  surface  is  1  sq.  yd. 

82.  If  a  circumference  is  described  on  the  surface  of  a  sphere  by  a 
pair  of  compasses  whose  points  are  2f  in.  apart,  what  is  the  area  of  the 
/one  bounded  by  this  circumference  ? 

83.  On  a  sphere  the  area  of  whose  surface  is  288  sq.  ft.  is  a  birectan- 
gular  spherical  triangle  whose  vertex  angle  is  100°.     Find  the  area  of 
this  triangle. 

84.  Five  inches  from  the  center  of  a  sphere  whose  diameter  is  two 
feet  a  plane  is  passed.     Find  the  areas  of  the  two  zones  formed.     Find 
the  chords  of  their  generating  arcs. 

85.  The  diameter  of  the  moon  is  about  2000  mi. ;  that  of  the  earth, 
about  8000  mi.     How  do  their  surfaces  compare?  their  volumes? 

86.  The  radii  of  two  concentric  spheres  are  12  in.  and  13  in.     A  plane 
is  tangent  to  the  inner  sphere.     Find  area  of  section  of  outer  sphere. 

87.  If  a  solid  sphere  4  ft.  in  diameter  weighs  500  lb.,  what  is  the 
weight  of  a  spherical  shell,  whose  external  diameter  is  10  ft.,  if  it  is  made 
of  the  same  material  and  a  foot  thick? 

88.  The  sun's  diameter  is  about  109  times  the  diameter  of  the  earth. 
How  do  the  areas  of  their  surfaces  compare?  their  volumes? 

89.  How  many  quarter-inch  spherical  bullets  can  be   made   from  a 
sphere  of  lead  a  foot  in  diameter? 

90.  A  12 -inch  cube  of  lead  is  melted  and  cast  in  the  form  of  a  spheri- 
cal cannon  ball.     What  is  the  radius  of  the  cannon  ball  ? 

91.  Find  the  angles  of  an  equiangular  spherical  triangle  equal  to  the 
sum  of   three  equiangular  spherical   triangles  (upon  the  same  sphere) 
whose  angles  are  each  75°. 

92.  What  is  the  radius  of  a  sphere  equal  to  the  sum  of  two  spheres 
whose  radii  are  3  in.  and  4  in.  respectively? 

93.  What  is  the  radius  of   a  sphere  equal  to  the  difference  of  two 
spheres  whose  radii  are  5  in.  and  4  in.  respectively? 

94.  The  area  of  an  equiangular  spherical  triangle  is  TT  sq.  in.,  and  the 
radius  of  the  sphere  is  4  in.     Find  the  angles  of  the  triangle. 

95.  The  volumes  of  two  spheres  are  to  each  other  as  64  :  343.     What 
is  the  ratio  of  their  surfaces  ? 

96.  Find  the  volumes  of  the  segments  of  a  sphere  whose  radius  is  12 
in.  formed  by  a  plane  whose  distance  from  the  center  is  9  in. 


450  BOOK   IX.     SOLID  GEOMETRY 

97.  If  the  radius  of  a  sphere  is  20  in.,  find : 

(a)  The  area  of  its  surface. 

(6)  The  area  of  a  zone  whose  altitude  is  2  in. 

(c)  The  edge  of  a  cube  inscribed  in  the  sphere. 

(J)  The  area  of  a  lune  whose  angle  is  80°. 

(e)  The  area  of  a  spherical  triangle  whose  angles  are  75°,  53°,  72°. 

(/)  The  area  of  a  spherical  polygon  whose  angles  are  68°,  119°, 

128°,  147°,  150°. 

(#)  The  area  of  a  birectangular  spherical  triangle  whose  vertex- 
angle  is  54°. 

(Ji)  The  area  of  a  zone  of  one  base  whose  altitude  is  5  in. 
(i)    The  radius  of  a  sphere  whose  surface  is  four  times  as  large. 
(/)  The  volume  of  the  sphere. 
(£)  The  volume  of  a  wedge  whose  angle  is  36°. 
(/)    The  volume  of  a  spherical  pyramid  whose  base  is  the  triangle 

of  exercise  (e). 
(in)  The  volume  of  the  spherical  sector  whose  base  is  the  zone  of 

exercise  (6). 
(n)  The  volume  of  the  spherical  cone  whose  base  is  the  zone  of 

exercise  (h). 
(0)  The  volume  of  a  spherical  segment  of  one  base,  whose  altitude 

is  6  in. 
(jo)  The  volume  of  a  spherical  segment  whose  altitude  is  4  in.  and 

the  radii  of  whose  bases  are  12  in.  and  16  in. 
(q)  The  radius  of  a  sphere  whose  volume  is  64  times  as  large. 

98.  The  angles  of  a  spherical  triangle  are  80°,  90°,  100°.     Find  the 
angle  of  an  equal  lune. 

99.  In  a  sphere  whose  radius  is  26  in.  two  parallel  planes  are  passed 
34  in.  apart.     The  radii  of  the  two  sections  are  10  in.  and  24  in.     Find 
the  volume  of  the  spherical  segment  included  between  the  planes. 

100.  In  a  certain  refrigerating  plant  is  a  large  tank  of  ice  water.     From 
this  tank  to  a  faucet  is  a  pipe  £  in.  inside  diameter,  and  42  ft.  long.     The 
faucet  is  opened  and  1  qt.  of  water  runs  out  every  4  seconds.     In  what 
length  of  time  will  the  cold  water  from  the  tank  appear  at  the  faucet? 

101.  A  sphere  2  ft.  in  diameter  is  trisected  by  two  concentric  spherical 
surfaces.     Find  the  radii  of  these  surfaces,  in  inches. 

102.  If  a  cylindrical  leaden  bar,  a  ft.  long  and  b  in.  in  diameter,  is 
melted  and  made  into  bullets,  |  in.  in  diameter,  explain  the  successive 
steps  necessary  to  be  taken  to  ascertain  the  number  of  bullets  there  will  be. 


SUMMARY   OF  FORMULAS   OF  SOLID   GEOMETRY 

B  —  area  of  base.  m  =  radius  of  mid-section. 

b  =  area  of  upper  base.  P  =  perimeter  of  base. 

E  =  number  of  edges.  Pr  =  perimeter  of  right  section, 

e,  e'  =  homologous  edges.  p  =  perimeter  of  upper  base. 

F  =  number  of  faces.  R,  r  —  radius  of  base. 

H  =  altitude.  s  =  slant  height. 

L  =  lateral  area.  T  =  total  area. 

M  =  mid-section.  V  =  volume  ;  number  of  vertices. 

PRISMS  AND  PYRAMIDS 

Parallelepiped V  =  BH  (584). 

Prism L  =  H   Pr  (569). 

T=L  +  2J3  (570). 

V=B    H  (590). 

Prismatoid V  =  \H(b  +  B  +  4  M)  (628). 

Regular  Pyramid L  =  *!»  •  s  (604). 

T=L  +  B  (595). 

Pyramid V  =  \B    H  (612). 

Frustum  of  pyramid     ....    L  =  ±(P  +  p~)s  (605). 

V  =  I  H(B  +  b  +  VB^b)  (618). 

Polyhedron.     .....      JEJ  +  2  =  F  4- F  (624). 

Sum  of  face  A  =  (V-  2)360°  (626). 

Similar  polyhedrons      .     .       T  :T  =  e*.-.e'*  (632). 

F:F'  =  e3:e'3  (634). 

CYLINDERS  AND  CONES 

Right  circular  cylinder  .     .     .     L  =  2TrRH  (654). 

T=2TTR(H+R)  (654). 

Circular  cylinder V=irR*H  (656). 

Right  circular  cone  ....  L  =  TTRs  (679). 

T  =  irR(s  +  R)  (679). 

Circular  cone V  =  \TtR*H  (683). 

• 

Frustum  of  right  circular  cone     L  =  TT(R  +  r)s  (681), 

=  2irms  (681). 

T  =  ir[(R  +  r)s  +  -R2  +  f2]  (681). 

V  =  l-  irJff  [  J?2  +  r2  +  B  •  r]  (685) . 

3 

EOBBINS'S    NEW   SOLID   GEOM.  —  14         451 


452 


BOOK  IX.     SOLID   GEOMETRY 


A  =  angle  of  lime. 

E  —  spherical  excess. 
H  =  altitude. 
L  =  area-  of  lime. 
R  =  radius  of  sphere. 


r,  r'  =  radii  of   bases  of   spherical 

segment. 

S  =  area  of  spherical  surface. 
V  =  volume. 
Z  =  area  of  zone. 
A  =  area  of  triangle. 


Spherical  surface    . 

THE  SPHERE 

.     .     S  =  4irR2 
.    .     Z  -  2  TTllH 

(774). 

(778). 

Zone  of  one  base     .     . 

.    .     Z  -  TT  (chord)2 
.    .     L  —  2  A  spherical  degrees 

(779). 
(781). 

or  L  —      '     ~^   souare  units 

<78°^ 

Spherical  triangle   .     . 

90 
.     .     A  =  E  spherical  degrees 

or  A  ~        Tf-R  square  units 

V.1  ^>")- 

(784). 
(785) 

Spherical  polygon  .     . 
Sphere       

ISO 
Polygon  =  E  spherical  degrees 
v     4irU3 

(780). 
(787). 

Spherical  pyramid 

3 

.    .     F  =  |(base)U 

r_^A-TTR* 

(789). 
(79l\ 

Spherical  sector      .     . 

.   .    V^ITTR^H 

.    .     F-firl?2JJ 

(793). 
(793). 

Spherical  segment,  one 
Any  spherical  segment 

base,     r  =  lTTH^R-H) 

F  =  -  TrH  (r*  +  r'2)  +  i  IT  IT3 

2                                        6 

(794). 
(794). 

INDEX   OF   DEFINITIONS 


(The  numbers  refer  to  pages.) 


Abbreviations,  vii. 

Adjacent  dihedral  angles,  290. 

Alternate    interior    dihedral    angles, 

291. 
Altitude,  of  cone,  377. 

of  cylinder.  367. 

of  frustum  of  cone,  378. 

of  frustum  of  pyramid,  333. 

of  prism,  314. 

of  prismatoid,  352. 

of  pyramid,  332. 

of  spherical  segment,  429. 

of  zone,  428. 

Angle,    between   intersecting  curves, 
395. 

birectangular  trihedral,  307. 

convex  polyhedral,  305. 

dihedral,  290. 

isosceles  trihedral,  307. 

plane  angle  of  dihedral,  291. 

polyhedral,  305. 

rectangular  trihedral,  307. 

spherical,  395. 

trihedral,  307. 

trirectangular  trihedral,  307. 
Angles,  adjacent  dihedral,  290. 

complementary  dihedral,  291. 

equal  dihedral,  291. 

equal  polyhedral,  306. 

face,  of  polyhedral,  305. 

of  lune,  427. 

of  spherical  triangle,  408. 

right  dihedral,  291. 

supplementary  dihedral,  291. 

symmetrical  polyhedral,  306. 

vertical  dihedral,  291. 

vertical  polyhedral,  306. 


Axis,  of  circle  of  sphere,  394. 
of  circular  cone,  377. 

Base,  of  cone,  377. 

of  pyramid,  332. 

of  spherical  pyramid,  428. 

of  spherical  sector,  428. 
Bases,  of  cylinder,  367. 

of  prism,  314. 

of  spherical  segment,  429. 

of  zone,  428. 

Birectangular  spherical  triangle,  408. 
Birectangular  trihedral  angle,  307. 

Center  of  sphere,  393. 
Circles  of  sphere,  394. 

line  tangent  to,  395. 
Circular  cone,  377. 

axis  of,  377. 

right,  377. 
Circular  cylinder,  368. 

right,  368. 
Circumscribed  frustum   of   pyramid, 

378. 

Circumscribed  prism,  368. 
Circumscribed  sphere   about  polyhe- 
dron, 395. 

Complementary  dihedral  angles,  291. 
Cone,  377. 

altitude  of,  377. 

base  of,  377. 

circular,  377. 

circular,  axis  of,  377. 

frustum  of,  378. 

lateral  area  of,  377. 

lateral  area  of  frustum  of,  378. 

oblique  circular,  377. 


453 


454 


INDEX   OF   DEFINITIONS 


Cone —  (continued] 

of  revolution,  378. 

of  revolution,  slant,  height  of,  378. 

plane  tangent  to,  378. 

right  circular,  377. 

spherical,  428. 

total  area  of,  377. 

total  area  of  frustum  of,  378. 
Cones,  377. 

similar,  of  revolution,  378. 
Congruent  solids,  315. 
Congruent  spherical  polygons,  400. 
Conical  surface,  377. 
Convex  polyhedral  angle,  305. 
Convex  polyhedron,  313. 
Corresponding  dihedral  angles,  201. 
Cube,  315,  354. 
Cylinder,  367. 

altitude  of,  367. 

bases  of,  367. 

circular,  368. 

lateral  area  of,  367. 

oblique,  368. 

of  revolution,  368. 

right,  368. 

right  circular,  368. 

right  section  of,  368. 

total  area  of,  367. 
Cylinders,  367. 

similar,  of  revolution,  368. 
Cylindrical  surface,  367. 

Degree,  spherical,  428. 
Determined,  plane,  262. 
Diagonal,  of  polyhedron,  313. 

of  spherical  polygon,  400. 
Diameter  of  sphere,  303. 
Dihedral  angle,  200. 

edge  of,  200. 

faces  of,  200. 

plane  angle  of,  201. 
Dihedral  angles,  200. 

adjacent,  200. 

alternate  interior,  201. 

complementary,  201. 

corresponding,  201. 

equal,  201. 


Dihedral  angles —  (continued) 

right,  201. 

supplementary,  201. 

vertical,  201. 

Dimensions  of  parallelepiped,  315. 
Directrix,  367,  377. 
Distance,  between  points  of  surface  of 
sphere,  305. 

from  point  to  a  plane,  286. 
Dodecahedron,  313. 

regular,  354. 

Edge  of  dihedral  angle,  200. 
Edges,  of  polyhedral  angle,  305. 

of  polyhedron,  313. 
Element,  of  conical  surface,  377. 

of  cylindrical  surface,  367. 
Equal  dihedral  angles,  201. 
Equal  polyhedral  angles,  306. 
Equal  solids,  315. 
Equal  spheres,  305. 
Equiangular  spherical  triangles,  408. 
Equilateral  spherical  triangles,  408. 
Excess,  spherical,  428. 

Face  angles  of  polyhedral  angle,  305. 
Faces,  of  dihedral  angle,  200. 

of  polyhedral  angle,  305. 

of  polyhedron,  313. 
Foot  of  line,  261. 
Frustum  of  cone,  378,. 

altitude  of,  378. 

lateral  area  of,  378. 

mid-section  of,  378. 

slant  height  of,  378. 

total  area  of,  378. 
Frustum  of  pyramid,  333. 

altitude  of,  333. 

circumscribed    about    frustum    of 
cone,  378. 

inscribed  in  frustum  of  cone,  378. 

slant  height  of,  333. 

Generatrix,  367,  377. 
Geometrical  solid,  316. 
Geometry,  Solid,  261. 
Great  circle  of  sphere,  394. 


INDEX   OF   DEFINITIONS 


455 


Great  circle,  axis  of,  394. 

Hemisphere,  429. 
Hexahedron,  313. 

regular,  354. 
Historical  notes : 

Archimedes,  393. 

August,  353. 

Eudoxus,  345. 

Gerard,  410. 

Hippasus,  355. 

Menelaus,  393. 

Pythagoras,  355. 

Icosahedron,  313. 

regular,  354. 
Inclination  of  line,  286. 
Inscribed  frustum  of  pyramid,  378. 
Inscribed  prism,  338. 
Inscribed  prism  in  cylinder,  368. 
Inscribed  sphere  in  polyhedron,  395. 
Intersection,  261. 
Isosceles  trihedral  angle,  307. 

Lateral  area,  of  cone,  377. 

of  cylinder,  367. 

of  frustum  of  cone,  378. 

of  prism,  314. 

of  pyramid,  332. 
Lateral  edges,  of  prism,  314. 

of  pyramid,  332. 
Lateral  faces,  of  prism,  314. 

of  pyramid,  332. 
Line,  foot  of,  261. 

inclination  of,  286. 

projection  of,  262. 

straight,  oblique  to  plane,  261. 

straight,  parallel  to  plane,  261. 

straight,  perpendicular  to  plane,  261. 

tangent  to  circle  of  sphere,  395. 

tangent  to  sphere,  395. 
Lune,  427. 

angles  of,  427. 

vertices  of,  427. 

Mid-section  of  frustum  of  cone,  378. 
of  prismatoid,  352. 


Mutually  equiangular  spherical  tri- 
angles, 408. 

Mutually  equilateral  spherical  tri- 
angles, 408. 

Normal,  261. 

Oblique  circular  cone,  377. 
Oblique  cylinder,  368. 
Oblique  parallelepiped,  315. 
Oblique  prism,  314. 
Octahedron,  313. 
regular,  354. 

Parallel  planes,  261. 
Parallelepiped,  315. 

dimensions  of,  316. 

oblique,  315. 

rectangular,  315. 

right,  315. 

Perpendicular  planes,  291. 
Plane,  261. 

determined,  262. 

distance  from  point  to,  286. 

straight  line  oblique  to,  261. 

straight  line  parallel  to,  261. 

straight  line  perpendicular  to,  261. 

tangent  to  cone,  378. 

tangent  to  cylinder,  368. 

tangent  to  sphere,  395. 
Plane  angle  of  dihedral  angle,  291. 
Plane  section    of    polyhedral    angle, 

305. 
Planes,  parallel,  261. 

perpendicular,  291. 
Point,  of  contact,  395. 

of  tangency,  395. 

projection  of,  262. 
Polar  distance,  395. 
Polar  triangle,  409. 
Poles  of  circle  of  sphere,  394. 
Polygon,  spherical,  409. 
Polyhedral  angle,  305. 

convex,  305. 

edges  of,  305. 

face  angles  of,  305. 

faces  of,  305. 


456 


INDEX   OF   DEFINITIONS 


Polyhedral  angle  —  (continued) 

plane  section  of,  305. 

vertex  of,  305. 
Polyhedral  angles,  306. 

equal,  306. 

symmetrical,  306. 

vertical,  306. 
Polyhedron,  313. 

convex,  313. 

diagonal  of,  313. 

edges  of,  313. 

faces  of,  313. 

inscribed  sphere  in,  395. 

regular,  353. 

vertices  of,  313. 
Polyhedrons,  353. 

similar,  353. 
Prism,  314. 

altitude  of,  314. 

bases  of,  314. 

circumscribed  about  cylinder,  368. 

circumscribed  about  pyramid,  338. 

inscribed  in  cylinder,  368. 

inscribed  in  pyramid,  338. 

lateral  area  of,  314. 

lateral  edges  of,  314. 

lateral  faces  of,  314. 

oblique,  314. 

regular,  314. 

right,  314. 

right  section  of,  315. 

total  area  of,  314. 

triangular,  314. 

truncated,  314. 
Prismatoid,  351. 

altitude  of,  352. 

mid-section  of,  352. 
Prismoid,  352. 
Projection,'262. 
Pyramid,  332. 

altitude  of,  332. 

altitude  of  frustum  of,  333. 

base  of,  332. 

circumscribed  about  cone,  378. 

circumscribed  frustum  of,  378. 

frustum  of,  333. 

inscribed  frustum  of,  378. 


Pyramid —  (continued} 

inscribed  in  cone,  378. 

lateral  area  of,  332. 

lateral  edges  of,  332. 

lateral  faces  of,  332. 

regular,  332. 

slant  height  of  frustum  of  regular, 
333. 

slant  height  of  regular,  333. 

spherical,  428. 

total  area  of,  332. 

triangular,  332. 

truncated,  333. 

vertex  of,  332. 
Pyramids,  332. 

Quadrant,  395. 

Radius  of  sphere,  393. 
Rectangular  parallelepiped,  315. 
Rectangular  trihedral  angle,  307. 
Regular  dodecahedron,  354. 
Regular  hexahedron,  354. 
Regular  icosahedron,  354. 
Regular  octahedron,  354. 
Regular  polyhedron,  353,  354. 
Regular  prism,  314. 
Regular  pyramid,  332. 

slant  height  of,  333. 
Regular  tetrahedron,  354. 
Revolution,  cone  of,  378. 

cylinder  of,  368. 

similar  cones  of,  378. 

similar  cylinders  of,  368. 
Right  circular  cone,  377. 
Right  circular  cylinder,  368. 
Right  cylinder,  368. 
Right  dihedral  angles,  21)1. 
Right  parallelepiped,  315. 
Right  prism,  314. 
Right  section,  of  cylinder,  368. 

of  prism,  315. 

Sector,  base  of  spherical,  428. 

spherical,  428. 
Segment,  base  of  spherical,  429. 

spherical,  428. 


INDEX   OF   DEFINITIONS 


457 


Sides  of  spherical  triangle,  408. 
Similar  cones  of  revolution,  378. 
Similar  cylinders  of  revolution,  368. 
Similar  polyhedrons,  353. 
Slant  height,  of  cone  of  revolution, 
378. 

of  frustum  of  cone,  378. 

of  frustum  of  pyramid,  333. 

of  regular  pyramid,  333. 
Small  circle  of  sphere,  394. 
Solid,  261. 

geometrical,  316. 

volume  of,  315. 
Solid  geometry,  261. 
Solids,  congruent,  315. 

equal,  315. 
Sphere,  393. 

axis  of  circle  of,  394. 

center  of,  393. 

circumscribed    about    polyhedron, 
395. 

diameter  of,  393. 

great  circle  of,  394. 

inscribed  in  polyhedron,  395. 

line  tangent  to,  395. 

line  tangent  to  circle  of,  395. 

plane  tangent  to,  395. 

small  circle  of,  394. 
Spheres,  equal,  395. 

tangent,  395. 
Spherical  angle,  395. 
Spherical  cone,  428. 
Spherical  degree,  428. 
Spherical  excess,  428. 
Spherical  polygon,  409. 

diagonal  of,  409. 

Spherical  polygons,  congruent,  409. 
Spherical  pyramid,  428. 

base  of,  428. 

vertex  of,  428. 
Spherical  sector,  428. 

base  of,  428. 
Spherical  segment,  428. 

altitude  of,  429. 

bases  of,  429. 

of  one  base,  429. 
Spherical  surface,  393. 


Spherical  triangle,  408. 

angles  of,  408. 

birectangular,  408. 

sides  of,  408. 

symmetrical,  409. 

trirectangular,  408. 

unit  of  measure  of,  408. 

vertices  of,  408. 
Spherical  triangles,  408. 

mutually  equiangular,  408. 

mutually  equilateral,  408. 
Spherical  wedge,  429. 
Straight    line,     oblique     to     plane, 
261. 

parallel  to  plane,  261. 

perpendicular  to  plane,  261. 
Supplementary  dihedral  angles,  291. 
Surface,  conical,  377. 

cylindrical,  367. 

spherical,  393. 
Surfaces,  261. 
Symmetrical       polyhedral       angles, 

306. 

Symmetrical      spherical      triangles, 
409. 

Tangent  spheres,  395. 
Tetrahedron,  313. 

regular,  354. 
Total  area,  of  cone,  377. 

of  cylinder,  367. 

of  frustum  of  cone,  378. 

of  prism,  314. 

of  pyramid,  332. 
Triangle,  spherical,  408. 
Triangular  prisrn,  314. 
Triangular  pyramid,  332. 
Trihedral  angle,  307. 

birectangular,  307. 

isosceles,  307. 

rectangular,  307. 

trirectangular,  307. 
Trirectangular      spherical      triangle, 

408. 

Trirectangular  trihedral  angle,  307. 
Truncated  prism,  314. 
Truncated  pyramid,  333. 


458 


INDEX   OF   DEFINITIONS 


Unit,  of  measure  of  spherical  triangle, 

408. 
of  volume,  315. 

Vertex,  of  polyhedral  angle,  305. 

of  pyramid,  332. 

of  spherical  pyramid,  428. 
Vertical  dihedral  angles,  291. 
Vertical  polyhedral  angles,  306. 
Vertices,  of  lune,  427. 


Vertices,  of  polyhedron,  313. 

of  spherical  triangle,  408. 
Volume,  of  solid,  315. 

unit  of,  315. 

Wedge,  spherical,  429. 

Zone,  428. 
altitude  of,  428. 
bases  of,  428. 
of  one  base,  428. 


ADVERTISEMENTS 


ROBBINS'S     PLANE 
TRIGONOMETRY 

By   EDWARD   R.    ROBBINS,  Senior  Mathematical  Mas- 
ter, William  Perm  Charter  School,  Philadelphia,  Pa. 

$0.60 


THIS  book  is  intended  for  beginners.      It  aims  to  give  a 
thorough   familiarity   with    the   essential   truths,    and    a 
satisfactory  skill  in  operating  with  those  processes.      It 
is  illustrated  in  the  usual  manner,  but  the  diagrams  are  more 
than  usually  clear-cut  and  elucidating. 

^j  The  work  is  sound  and  teachable,  and  is  written  in  clear 
and  concise  language,  in  a  style  that  makes  it  easily  under- 
stood. Immediately  after  each  principle  has  been  proved, 
it  is  applied  first  in  illustrative  examples,  and  then  further  im- 
pressed by  numerous  exercises.  Accuracy  and  rigor  of  treat- 
ment are  shown  in  every  detail,  and  all  irrelevant  and  ex- 
traneous matter  is  excluded,  thus  giving  greater  prominence 
to  universal  rules  and  formulas. 

^|  The  references  to  Plane  Geometry  preceding  the  first 
chapter  are  invaluable.  A  knowledge  of  the  principles  of 
geometry  needed  in  trigonometry  is,  as  a  rule,  too  freely  taken 
for  granted.  The  author  gives  at  the  beginning  of  the  book 
a  statement  of  the  applied  principles,  with  reference  to  the 
sections  of  his  Geometry,  where  such  theorems  are  proved  in 
full.  Cross  references  in  the  text  of  the  Trigonometry  to 
those  theorems  make  it  easy  for  the  pupil  to  review  or  to 
supplement  imperfect  knowledge. 

^[  Due  emphasis  is  given  to  the  theoretical  as  well  as  to  the 
practical  applications  of  the  science.  The  number  of  ex- 
amples, both  concrete  and  abstract,  is  far  in  excess  of  those 
in  other  books  on  the  market.  This  book  contains  four  times 
as  many  exercises  as  most  books,  and  twice  as  many  as  that 
having  the  next  lowest  number. 


AMERICAN     BOOK     COMPANY 

(313) 


MILNE'S       STANDARD 
ALGEBRA 

By  WILLIAM  J.    MILNE,  Ph.D.,  LL.D.,   President  of 
the  New  York  State  Normal  College,  Albany,  N.  Y. 

$1.00 


THE  Standard  Algebra  conforms  to  the  most  recent 
courses  of  study.  The  inductive  method  of  presentation 
is  followed,  but  declarative  statements  and  observations 
are  used,  instead  of  questions.  Added  to  this  kind  of  unfold- 
ing and  development  of  the  subject  are  illustrative  problems 
and  explanations  to  bring  out  specific  points,  the  whole  being 
driven  home  by  varied  and  abundant  practice, 
^f  The  problems  are  fresh  in  character,  and  besides  the  tradi- 
tional problems  include  a  large  number  drawn  from  physics, 
geometry,  and  commercial  life.  They  are  classified  accord- 
ing to  the  nature  of  the  equations  involved,  not  according  to 
subject  matter.  The  statement  of  necessary  definitions  and  of 
principles  is  clear  and  concise,  but  the  proofs  of  principles, 
except  some  important  ones,  are  left  for  the  maturer  years  of 
the  pupil. 

^f  Accuracy  and  self-reliance  are  encouraged  by  the  use  of 
numerous  checks  and  tests,  and  by  the  requirement  that  re- 
sults be  verified.  The  subject  of  graphs  is  treated  after  simple 
equations,  introduced  by  some  of  their  simple  uses  in  repre- 
senting statistics,  and  in  picturing  two  related  quantities  in 
the  process  of  change,  and  again  after  quadratics.  Later  they 
are  utilized  in  discussing  the  values  of  quadratic  expressions. 
Factoring  receives  particular  attention.  Not  only  are  the 
usual  cases  given  fully  and  completely  with  plenty  of  practice, 
but  the  factor  theorem  is  taught. 

^f  The  helpful  and  frequent  reviews  are  made  up  of  pointed 
oral  questions,  abstract  exercises,  problems,  and  recent  college 
entrance  examination  questions.  The  book  is  unusually  handy 
in  size  and  convenient  for  the  pocket.  The  page  size  is  small. 


AMERICAN    BOOK    COMPANY 


ADVANCED     ARITHMETIC 

$0.75 

By    ELMER    A.     LYMAN,    Professor    of    Mathematics, 
Michigan  State  Normal  College,  Ypsilanti 


THIS   book  meets  the  requirements   of  secondary  and 
normal  schools.     In  its  preparation  the  author  has  aimed 
to  make  the  work  a  study  of  the  fundamental  principles 
of  arithmetic,  and  thereby  emphasize  the  disciplinary  value  of 
the  subject,  and  at  the  same  time  to  apply  these  principles  to 
the  solution  of  practical  business  problems.      To  this  end  such 
methods  as  are  used  in  the  best  commercial  practice  are  em- 
phasized throughout    the  work,    and   obsolete   methods  and 
problems  are  carefully  excluded. 

^[  The  exercises  have  been  selected  largely  from  actual  busi- 
ness transactions,  and  nearly  all  of  the  problems  in  the  appli- 
cation of  percentage  have  been  secured  from  business  houses, 
or  reviewed  by  representative  business  men.  The  chapters 
on  banking,  and  stocks  and  bonds,  give  information  of  a  prac- 
tical character  which,  though  indispensable  to  a  proper  under- 
standing of  the  subject,  is  rarely  found  in  text-books. 
^j  In  order  to  economize  time,  pupils  are  encouraged  to  use 
every  practical  labor-saving  device  known  to  the  science  of 
arithmetic,  but  so-called  short  processes,  which  are  compli- 
cated or  cumbersome,  have  been  carefully  avoided.  The  use 
of  checks  is  also  strongly  recommended,  because  it  contributes 
greatly  to  accuracy  in  results,  and  cultivates  a  spirit  of  self- 
reliance. 

^[  In  addition  to  the  special  methods  for  solution  given  in 
connection  with  the  various  subjects,  a  chapter  is  devoted  to 
the  general  method  of  approach  to  any  problem.  This  offers 
pupils  much  helpful  advice  in  attempting  the  solution  of  prob- 
lems of  a  miscellaneous  character,  such  as  are  given  in  exam- 
inations. Attention  is  also  called  to  the  historical  notes,  to 
the  treatment  of  graphical  representations,  and  to  the  chapter 
on  approximate  results. 


AMERICAN    BOOK     COMPANY 

(58) 


PLANE  SURVEYING  FOR  USE 

IN  THE   CLASSROOM  AND 

FIELD 

$3.00 

By  WILLIAM  G.  RAYMOND,  C.E.,  LL.D.,  Member 
American  Society  of  Civil  Engineers,  Dean  of  the  Col- 
lege of  Applied  Science,  State  University  of  Iowa. 


THIS  standard  textbook  has  now  been  completely  re- 
vised, rewritten,  rearranged,  reset  and  remade.      The 
new  edition  is  distinguished  by  the  convenience  of  its 
pocket  form  for  field  service ;   the  completeness  of  its  text ; 
and  the  clearness  of  its  tables.      The  book  is  light  in  weight, 
being  printed  on  thin  Bible  paper,  and  is  bound  in  flexible 
leather  covers  with  rounded  corners,  so  that  it  can  be  rolled 
without  injury.      The  maps  have  been  especially  planned  for 
hard  service  and  therefore  include  no  folded  maps  that  are 
easily  torn. 

^[  The  principles  are  carefully  explained  and  the  exercises 
are  designed  to  show  the  student,  by  his  own  experience, 
not  only  the  possibilities,  but  also  the  limitations  of  instru- 
ments, methods,  and  individuals.  Among  the  improvements 
are:  a  new  and  very  practical  chapter  on  finding  the  merid- 
ian, latitude,  and  time;  a  practical  discussion  of  errors; 
eighty  new  cuts  and  a  colored  frontispiece ;  more  than  twice 
the  number  of  exercises  and  examples ;  a  new  discussion  of 
the  adjustments  of  instruments,  with  what  is  believed  to  be 
the  first  technically  correct  statement  of  the  adjustment  of  the 
line  of  sight  of  the  transit ;  important  additions  to  the  chapters 
on  city  surveying,  railroad  curves,  earthwork,  and  hydro- 
graphic  surveying;  and  a  more  systematic  order  of  treatment 
and  arrangement.  The  Crockett  logarithmic  tables,  which 
are  the  best  for  the  use  of  surveyors,  have  been  expressly 
made  for  this  book  and  are  still  further  improved. 


AMERICAN     BOOK     COMPANY 

(76) 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 
£*U  CATION  LIBRA**  

AN  INITIAL  FINE  OF  25  CENTS 

WILL   BE   ASSESSED    FOR    FAILURE  TO    RETURN 
THIS    BOOK    ON    THE    DATE    DUE.    THE    PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY     AND     TO    $1.OO    ON    THE    SEVENTH     DAY 
OVERDUE. 

JsTtO 

LD  21-100»i-12,'43(879Gs) 

JU  4V584 


459960 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


